Mẹo What is the smallest number that 1225 must be multiplied by to make it a perfect cube? ?
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Question 1.
Nội dung chính Show- What should be multiplied by 1225 to make a perfect cube?What is the smallest number that should be multiplied to make it a perfect cube?Is 7803 a perfect cube?Is 10976 a perfect cube?
Let's find perfect cubes from the following numbers
125, 500, 64, 73, 729, 968
Answer:
Let us understand, what perfect cube is.
A perfect cube is a number that is a cube of an integer.
So,
To find the perfect cube from the following numbers, we need to find their prime factors individually and then group them into triplets of the prime factors.
If no prime factor is left out, then the number is a perfect cube.
However, if one of the prime factors is a single factor or double factor, then the number is not a perfect cube.
Take 125.
Prime factors of 125 = 5 × 5 × 5
Group them into triplets.
⇒ Prime factors of 125 = (5 × 5 × 5)
Since, prime factor 5 form a triplet, so 125 is a perfect cube.
∴, 125 is a perfect cube.
Take 500.
Prime factors of 500 = 5 × 5 × 5 × 2 × 2
Group the factors into triplets.
⇒ Prime factors of 500 = (5 × 5 × 5) × 2 × 2
Since, prime factor 2 is not part of a triplet, so 500 is not a perfect cube.
∴, 500 is not a perfect cube.
Take 64.
Prime factors of 64 = 2 × 2 × 2 × 2 × 2 × 2
Group the factors into triplets.
⇒ Prime factors of 64 = (2 × 2 × 2) × (2 × 2 × 2)
Since, prime factor 2 forms two triplets, so 64 is a perfect cube.
∴, 64 is a perfect cube.
Take 73.
Prime factors of 73 = 7 × 7 × 7
Group the factors into triplets.
⇒ Prime factors of 73 = (7 × 7 × 7)
Since, prime factor 7 forms a triplets, so 73 is a perfect cube.
∴, 73 is a perfect cube.
Take 729.
Prime factors of 729 = 3 × 3 × 3 × 3 × 3 × 3
Group the factors into triplets.
⇒ Prime factors of 729 = (3 × 3 × 3) × (3 × 3 × 3)
Since, prime factor 3 forms two triplets, so 729 is a perfect cube.
∴, 729 is a perfect cube.
Take 968.
Prime factors of 968 = 2 × 2 × 2 × 11 × 11
Group the factors into triplets.
⇒ Prime factors of 968 = (2 × 2 × 2) × 11 × 11
Since, prime factor 11 is not a part of a triplet, so 968 is not a perfect cube.
∴, 968 is not a perfect cube.
Lets Work Out 5.1
Question 1.
Let’s form two cubes with sides of length 5 cm and 1 cm
Let us calculate the number of small cube to form the big cube.
Answer:
We have to form two cubes of lengths 5 cm and 1 cm.
We know, a cube is symmetrical three-dimensional shape, either solid or hollow, contained by six equal squares.
We need to find the number of small cubes to form the big cube.
For this, we need to find the volume of the big cube.
Volume is the amount of space that a substance or object occupies, or that is enclosed within a container, and that is why we need to find volume to further find number of small cubes in that big cube.
Take the cube of length 5 cm.
Volume of cube of length 5 cm is given as
Volume = length × length × length
Since, length = 5 cm
⇒ Volume = 5 × 5 × 5 cm3
⇒ Volume = 125 cm3
This means, in the cube of length 5 cm, we have 125 cm cubes.
Now, take the cube of length 1 cm.
Volume of cube of length 1 cm is given as
Volume = length × length × length
Since, length = 1 cm
⇒ Volume = 1 × 1 × 1 cm3
⇒ Volume = 1 cm3
This means, in the cube of length 1 cm, we have 1 cm cubes.
Question 2.
Sumanta has made many cubes of 1 cm length. Manami is trying to make big cubes with those cubes. Let’s see in which of the cases given below Manami will be able to make big cubes :
i. 100 ii. 1000
iii. 1331 iv. 1210
v. 3375 vi. 2700
Answer:
Given that,
Sumanta has made many cubes of length, 1 cm.
This means, volume of each cube = length × length × length
Here, length = 1 cm
⇒ Volume of each cube = 1 × 1 × 1 cm3
⇒ Volume of each cube = 1 cm3
And a single cube out of those many cubes is made by Manami.
Let us assume that Sumanta has made ‘Z’ cubes of length 1 cm, where Z is some natural number.
So, if volume of 1 cube = 1 cm3
⇒ Volume of Z cubes = Z × 1 cm3
⇒ Volume of Z cubes = Z cm3
We have volumes of cube in the options.
(i). Let us check for 100.
We need to find cube root of 100. If we are able to find the cube root of 100, then we can say that a big cube is possible in this case.
100 = 2×2× 5× 5
We can see 100 cannot be written as a perfect cube.
∴ 100 ≠ (any integer)3
Since, we have cubes of length 1 cm each.
So, length of the final big cube will also be a natural number.
And cube root of 100 is not a natural number.
Thus, Manami won’t be able to make a big cube in this case.
(ii). Let us check for 1000.
We need to find the cube root of 1000. If we are able to find the cube root of 1000, then we can say that a big cube is possible in this case.
Cube root of 1000
1000 = 10× 10× 10
⇒ Cube root of 1000 = (1000)1/3
= 10
Since, we have cubes of length 1 cm each.
So, length of the final big cube will also be a natural number.
And cube root of 1000 is a natural number.
Thus, Manami will be able to make a big cube in this case.
(iii). Let us check for 1331.
We need to find the cube root of 1331. If we are able to find the cube root of 1331, then we can say that a big cube is possible in this case.
Cube root of 1331
1331 = 11× 11× 11
⇒ Cube root of 1331 = (1331)1/3
= 11
Since, we have cubes of length 1 cm each.
So, length of the final big cube will also be a natural number.
And cube root of 1331 is a natural number.
Thus, Manami will be able to make a big cube in this case.
(iv). Let us check for 1210.
We need to find the cube root of 1210. If we are able to find the cube root of 1210, then we can say that a big cube is possible in this case.
Cube root of 1210
1210 = 2× 5× 11× 11
We can see 1210 cannot be written as a perfect cube.
∴ 1210 ≠ (any integer)3
Since, we have cubes of length 1 cm each.
So, length of the final big cube will also be a natural number.
And cube root of 1210 is not a natural number.
Thus, Manami will not be able to make a big cube in this case.
(v). Let us check for 3375.
We need to find the cube root of 3375. If we are able to find the cube root of 3375, then we can say that a big cube is possible in this case.
Cube root of 3375
3375 = 15× 15× 15
⇒ Cube root of 3375 = (3375)1/3
= 15
Since, we have cubes of length 1 cm each.
So, length of the final big cube will also be a natural number.
And cube root of 3375 is a natural number.
Thus, Manami will be able to make a big cube in this case.
(vi). Let us check for 2700.
We need to find the cube root of 2700. If we are able to find the cube root of 2700, we can say that a big cube is possible in this case.
Cube root of 2700
2700 = 3× 3× 3× 2× 2× 5× 5
We can see 2700 cannot be written as a perfect cube.
∴ 2700 ≠ (any integer)3
Since, we have cubes of length 1 cm each.
So, length of the final big cube will also be a natural number.
And cube root of 2700 is not a natural number.
Thus, Manami will not be able to make a big cube in this case.
Question 3.
Let us write which number is not a perfect cube in the numbers given below.
i. 216 ii. 343
iii. 1024 iv. 324
v. 1744 vi. 1372
Answer:
Let us understand what perfect cube is.
A perfect cube is a number that is the cube of an integer.
(i). Let us check for 216.
We need to check whether 216 is a cube of an integer or not.
Let us factorize 216 for ease of calculation.
We have,
216 = 2 × 2 × 2 × 3 × 3 × 3
Group these factors into three similar integers.
216 = (2 × 2 × 2) × (3 × 3 × 3)
Take cube root on both sides,
Cube root of 216 is an integer.
Thus, 216 is a perfect cube.
(ii). Let us check for 343.
We need to find whether 343 is cube of an integer or not.
Let us factorize 343 for ease of calculation.
We have,
343 = 7 × 7 × 7
Group these factors into three similar integers.
343 = (7 × 7 × 7)
Take cube root on both sides,
Cube root of 343 is an integer.
Thus, 343 is a perfect cube.
(iii). Let us check for 1024.
We need to find whether 1024 is cube of an integer or not.
Let us factorize 1024 for ease of calculation.
We have,
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Group these factors into three similar integers.
1024 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 2
Take cube root on both sides,
Cube root of 1024 is clearly not an integer, since is not an integer.
Thus, 1024 is not a perfect cube.
(iv). Let us check for 324.
We need to find whether 324 is cube of an integer or not.
Let us factorize 324 for ease of calculation.
We have,
324 = 2 × 2 × 3 × 3 × 3 × 3
Group these factors into three similar integers.
324 = (3 × 3 × 3) × 2 × 2 × 3
Take cube root on both sides,
Cube root of 324 is clearly not an integer, since is not an integer.
Thus, 324 is not a perfect cube.
(v). Let us check for 1744.
We need to find whether 1744 is cube of an integer or not.
Let us factorize 1744 for ease of calculation.
We have,
1744 = 2 × 2 × 2 × 2 × 109
Group these factors into three similar integers.
1744 = (2 × 2 × 2) × 2 × 109
Take cube root on both sides,
Cube root of 1744 is clearly not an integer, since is not an integer.
Thus, 1744 is not a perfect cube.
(vi). Let us check for 1372.
We need to find whether 1372 is cube of an integer or not.
Let us factorize 1372 for ease of calculation.
We have,
1372 = 2 × 2 × 7 × 7 × 7
Group these factors into three similar integers.
1372 = (7 × 7 × 7) × 2 × 2
Take cube root on both sides,
Cube root of 1372 is clearly not an integer, since is not an integer.
Thus, 1372 is not a perfect cube.
Question 4.
Debnath has made a cuboid whose length, breadth and height are 4 cm, 3 cm and 2 cm respectively. Let’s see how many cuboids of this type will form a cube.
Answer:
Given that,
Debnath made a cuboid of dimension:
Length = 4 cm
Breadth = 3 cm
Height = 2 cm
Since, volume of cuboid = length × breadth × height
⇒ Volume of cuboid = 4 × 3 × 2
= 24
Therefore, volume of a cuboid made by Debnath is 24 cm3.
Factors of 24 are:
24 = 2 × 2 × 2 × 3
This 24 cm3 is volume of one cuboid, which is not even perfect cube.
And to form a cube, we need to have perfect cube volumes.
So make 24 into a perfect cube.
Notice the factors of 24,
24 = 2 × 2 × 2 × 3
Group three similar integers, we get
24 = (2 × 2 × 2) × 3
Multiply both sides by (3 × 3), we get
24 × (3 × 3) = (2 × 2 × 2) × 3 × (3 × 3)
⇒ 216 = (2 × 2 × 2) × (3 × 3 × 3)
Number of cuboids = (3 × 3)
Because multiplying (3 × 3) makes a cube by the same number of cuboids.
⇒ Number of cuboids = 9
Thus, number of required cuboids are 9.
Question 5.
Let us calculate with which positive smallest number should be multiplied with the numbers below so that the product will be a perfect cube.
i. 675 ii. 200
iii. 108 iv. 121
v. 1225
Answer:
Let us recall what a perfect cube is.
A perfect cube is a number that is the cube of an integer.
(i). For 675.
We need to factorize 675. We get,
675 = 3 × 3 × 3 × 5 × 5
Group these factors into three similar integers. We have
675 = (3 × 3 × 3) × 5 × 5
To make it into a perfect cube, multiply both sides by 5.
We have
675 × 5 = (3 × 3 × 3) × 5 × 5 × 5
Now, group these factors into three similar integers.
3375 = (3 × 3 × 3) × (5 × 5 × 5)
We are able to group it in three similar integers. This means, 3375 is a perfect cube.
Thus, smallest positive number that should be multiplied with 675 is 5.
(ii). For 200.
We need to factorize 200. We get,
200 = 2 × 2 × 2 × 5 × 5
Group these factors into three similar integers. We have
200 = (2 × 2 × 2) × 5 × 5
To make it into perfect cube, multiply both sides by 5.
We have
200 × 5 = (2 × 2 × 2) × 5 × 5 × 5
Now, group these factors into three similar integers.
1000 = (2 × 2 × 2) × (5 × 5 × 5)
We are able to group it in three similar integers. This means, 1000 is a perfect cube.
Thus, smallest positive number that should be multiplied with 200 is 5.
(iii). For 108.
We need to factorize 108. We get,
108 = 2 × 2 × 3 × 3 × 3
Group these factors into three similar integers. We have
108 = (3 × 3 × 3) × 2 × 2
To make it into perfect cube, multiply both sides by 2.
We have
108 × 2 = (3 × 3 × 3) × 2 × 2 × 2
Now, group these factors into three similar integers.
216 = (3 × 3 × 3) × (2 × 2 × 2)
We are able to group it in three similar integers. This means, 216 is a perfect cube.
Thus, smallest positive number should be multiplied with 108 is 2.
(iv). For 121.
We need to factorize 121. We get,
121 = 11 × 11
We can’t group it into three similar integers. So, multiply both sides by 11.
121 × 11 = 11 × 11 × 11
Now, group these factors into three similar integers.
1331 = (11 × 11 × 11)
Since, we are able to group it in three similar integers. This means, 1331 is a perfect cube.
Thus, smallest positive number that should be multiplied with 121 is 11.
(v). For 1225.
We need to factorize 1225. We get,
1225 = 5 × 5 × 7 × 7
We can’t group it into three similar integers. So, multiply both sides by (5 × 7).
1225 × 5 × 7 = 5 × 5 × 7 × 7 × (5 × 7)
Now, group these factors into three similar integers.
42875 = (5 × 5 × 5) × (7 × 7 × 7)
Since, we are able to group it in three similar integers. This means, 42875 is a perfect cube.
Thus, smallest positive number that should be multiplied with 42875 is (5 × 7), i.e., 35.
Question 6.
Let us calculate by which positive smallest number should be divided with the numbers below so that the quotient will be a perfect cube.
i. 7000 ii. 2662
iii. 4394 iv. 6750
v. 675
Answer:
Let us recall what a perfect cube is.
A perfect cube is a number that is the cube of an integer.
(i). For 7000.
We need to factorize 7000. We get,
7000 = 7 × 2 × 5 × 2 × 5 × 2 × 5
Group these factors into three similar integers.
7000 = (2 × 2 × 2) × (5 × 5 × 5) × 7
To make it into perfect cube, divide it by 7 on both sides.
⇒ 1000 = (2 × 2 × 2) × (5 × 5 × 5)
Since, we are able to group it in three similar integers. This means, 1000 is a perfect cube.
Thus, smallest positive number that should be divided with 7000 is 7.
(ii). For 2662.
We need to factorize 2662. We get,
2662 = 2 × 11 × 11 × 11
Group these factors into three similar integers.
2662 = (11 × 11 × 11) × 2
To make it into perfect cube, divide it by 2 on both sides.
⇒ 1331 = (11 × 11 × 11)
Since, we are able to group it in three similar integers. This means, 1331 is a perfect cube.
Thus, smallest positive number that should be divided with 2662 is 2.
(iii). For 4394.
We need to factorize 4394. We get,
4394 = 2 × 13 × 13 × 13
Group these factors into three similar integers.
4394 = (13 × 13 × 13) × 2
To make it into perfect cube, divide it by 2 on both sides.
⇒ 2197 = (13 × 13 × 13)
Since, we are able to group it in three similar integers. This means, 2197 is a perfect cube.
Thus, smallest positive number that should be divided with 4394 is 2.
(iv). For 6750.
We need to factorize 6750. We get,
6750 = 5 × 5 × 5 × 3 × 3 × 3 × 2
Group these factors into three similar integers.
6750 = (5 × 5 × 5) × (3 × 3 × 3) × 2
To make it into perfect cube, divide it by 2 on both sides.
⇒ 3375 = (5 × 5 × 5) × (3 × 3 × 3)
Since, we are able to group it in three similar integers. This means, 3375 is a perfect cube.
Thus, smallest positive number that should be divided with 6750 is 2.
(v). For 675.
We need to factorize 675. We get,
675 = 5 × 5 × 3 × 3 × 3
Group these factors into three similar integers.
675 = (3 × 3 × 3) × 5 × 5
To make it into perfect cube, divide it by (5 × 5) on both sides.
⇒ 27 = (3 × 3 × 3)
Since, we are able to group it in three similar integers. This means, 27 is a perfect cube.
Thus, smallest positive number that should be divided with 675 is (5 × 5), i.e., 25.
Question 7.
Let us write the numbers given below as product of prime factors and write cube roots of the numbers:
i. 512 ii. 1728
iii. 5832 iv. 15625
v. 10648
Answer:
Let us recall the definition of prime factors and cube roots.
Prime factors are any of the prime numbers that can be multiplied to give the original number.
The cube root of a number is a special value that, when used in a multiplication three times, gives that number.
(i). For 512:
512 can be divided by prime number 2, quotient is 256.
256 can be divided by prime number 2, quotient is 128.
128 can be divided by prime number 2, quotient is 64.
64 can be divided by prime number 2, quotient is 32.
32 can be divided by prime number 2, quotient is 16.
16 can be divided by prime number 2, quotient is 8.
8 can be divided by prime number 2, quotient is 4.
4 can be divided by prime number 2, quotient is 2.
2 can be divided by prime number 2, quotient is 1.
So, 512 can be expressed in product of prime factors as:
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Group these factors into three similar integers.
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)
Taking cube root on both sides, we get
Thus, cube root of 512 is 8.
(ii). For 1728.
1728 can be divided by prime number 2, quotient is 864.
864 can be divided by prime number 2, quotient is 432.
432 can be divided by prime number 2, quotient is 216.
216 can be divided by prime number 2, quotient is 108.
108 can be divided by prime number 2, quotient is 54.
54 can be divided by prime number 2, quotient is 27.
27 can be divided by prime number 3, quotient is 9.
9 can be divided by prime number 3, quotient is 3.
3 can be divided by prime number 3, quotient is 1.
So, 1728 can be expressed in product of prime factors as:
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
Group these factors into three similar integers.
1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)
Taking cube root on both sides, we get
Thus, cube root of 1728 is 12.
(iii). For 5832.
5832 can be divided by prime number 2, quotient is 2916.
2916 can be divided by prime number 2, quotient is 1458.
1458 can be divided by prime number 2, quotient is 729.
729 can be divided by prime number 3, quotient is 243.
243 can be divided by prime number 3, quotient is 81.
81 can be divided by prime number 3, quotient is 27.
27 can be divided by prime number 3, quotient is 9.
9 can be divided by prime number 3, quotient is 3.
3 can be divided by prime number 3, quotient is 1.
So, 5832 can be expressed in product of prime factors as:
5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Group these factors into three similar integers.
5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)
Taking cube root on both sides, we get
Thus, cube root of 5832 is 18.
(iv). For 15625.
15625 can be divided by prime number 5, quotient is 3125.
3125 can be divided by prime number 5, quotient is 625.
625 can be divided by prime number 5, quotient is 125.
125 can be divided by prime number 5, quotient is 25.
25 can be divided by prime number 5, quotient is 5.
5 can be divided by prime number 5, quotient is 1.
So, 15625 can be expressed in product of prime factors as:
15625 = 5 × 5 × 5 × 5 × 5 × 5
Group these factors into three similar integers.
15625 = (5 × 5 × 5) × (5 × 5 × 5)
Taking cube root on both sides, we get
Thus, cube root of 15625 is 25.
(v). For 10648.
10648 can be divided by prime number 2, quotient is 5324.
5324 can be divided by prime number 2, quotient is 2662.
2662 can be divided by prime number 2, quotient is 1331.
1331 can be divided by prime number 11, quotient is 121.
121 can be divided by prime number 11, quotient is 11.
11 can be divided by prime number 11, quotient is 1.
So, 10648 can be expressed in product of prime factors as:
10648 = 2 × 2 × 2 × 11 × 11 × 11
Group these factors into three similar integers.
10648 = (2 × 2 × 2) × (11 × 11 × 11)
Taking cube root on both sides, we get
Thus, cube root of 10648 is 22.
Lets Work Out 5.2
Question 1.
Answer:
Given is, lengths of a side of cubes.
We need to find volumes of these cubes.
Let us recall the formula of volume of cube.
Volume of cube = length × length × length
⇒ Volume of cube = (length)3
(i). Given is,
Length of cube (unit) = p2 + q2
Then,
Volume of cube = (p2 + q2)3
Recall the algebraic identity,
(a + b)3 = a3 + b3 + 3a2b + 3ab2
Put a = p2 and b = q2. We get
(p2 + q2)3 = (p2)3 + (q2)3 + 3(p2)2(q2) + 3(p2) (q2)2
⇒ (p2 + q2)3 = p6 + q6 + 3p4q2 + 3p2q4
⇒ (p2 + q2)3= p6 + q6 + 3p2q2 (p2 + q2)
= p6 + q6 + 3p4q2 + 3p2q4
Thus, volume of cube is p6 + q6 + 3p4q2 + 3p2q4cubic unit.
(ii). Given is,
Then,
Recall the algebraic identity,
(a + b)3 = a3 + b3 + 3a2b + 3ab2
Put and . We get
Thus, volume of cube is cubic unit.
(iii). Given is,
Length of cube (unit) = x2y – z2
Then,
Volume of cube = (x2y – z2)3
Recall the algebraic identity,
(a – b)3 = a3 – b3 – 3a2b + 3ab2
Put a = x2y and b = z2. We get
(x2y – z2)3 = (x2y)3 – (z2)3 – 3(x2y)2(z2) + 3(x2y)(z2)2
= x6y3 – z6 – 3x4y2z2 + 3x2yz4
= x6y3 – z6 – 3x2yz2 (x2y – z2)
= x6y3 – z6 – 3x4y2z2 + 3x2 y z4
Thus, volume of cube is x6y3 – z6 – 3x4y2z2 + 3x2 y z4 cubic unit.
(iv). Given is,
Length of cube (unit) = 1 + b – 2c
Then,
Volume of cube = (1 + b – 2c)3
Recall the algebraic identity,
(a + b + c)3 = a3 + b3 + c3 + 3(a + b) (b + c)(c + a)
Put a = 1, b = b and c = -2c. We get
(1 + b – 2c)3 = (1)3 + (b)3 + (-2c)3 + 3(1 + b) (b – 2c)(-2c + 1)
⇒ (1 + b – 2c)3 = 1 + b3 – 8c3 + 3(1 + b) (b – 2c)(1 – 2c)
Thus, volume of cube is [1 + b3 – 8c3 + 3(1 + b)(b – 2c)(1 – 2c)] cubic unit.
(v). Given is,
Volume of cube (cubic unit) = (2.89)3 + (2.11)3 + 15 × 2.89 × 2.11
We need to find length of cube.
Volume of cube can be written as,
Volume = (2.89)3 + (2.11)3 + [3 × 5 × 2.89 × 2.11]
[∵, 15 = 3 × 5]
⇒ Volume = (2.89)3 + (2.11)3 + [3 × (2.89 + 2.11) × 2.89 × 2.11]
[∵, 5 = 2.89 + 2.11]
Put 2.89 = a and 2.11 = b. We get
Volume = a3 + b3 + [3 × (a + b) × a × b]
Or, Volume = a3 + b3 + 3ab (a + b)
Or, Volume = (a + b)3 …(a)
[∵, by algebraic identity, we know that
(a + b)3 = a3 + b3 + 3ab (a + b)]
Again,
Put a = 2.89 and b = 2.11. We get
Volume = (2.89 + 2.11)3
⇒ Volume = (5)3
We know, volume of cube is given by
Volume = (length)3
Or (length)3 = Volume
⇒ length = (Volume)1/3
⇒ length = 5
Thus, length of cube is 5 unit.
(vi). Given is,
Volume of cube (cubic unit) = (2m + 3n)3 + (2m – 3n)3 + 12m (4m2 – 9n2)
We need to find length of cube.
Volume of cube can be written as,
[∵, By identity, (a + b) (a – b) = a2 – b2 in (2m + 3n) (2m – 3n)]
Volume = (2m + 3n)3 + (2m – 3n)3 + [3 × 4m × (2m + 3n) (2m – 3n)]
[∵, 4m = 2m + 2m]
= (2m + 3n)3 + (2m – 3n)3 + [3 × (2m + 2m) × (2m + 3n) (2m – 3n)]
[∵, 3n – 3n = 0 and it won’t affect the equation]
⇒ Volume = (2m + 3n)3 + (2m – 3n)3 + [3 × (2m + 2m + 3n – 3n) × (2m + 3n) (2m – 3n)]
[∵, (2m + 2m + 3n – 3n) = (2m + 3n) + (2m – 3n); we have just rearranged]
⇒ Volume = (2m + 3n)3 + (2m – 3n)3 + [3 × ((2m + 3n) + (2m – 3n)) × (2m + 3n)(2m –
3n)]
⇒ Volume = (2m + 3n)3 + (2m – 3n)3 + 3[(2m + 3n) + (2m – 3n)][(2m + 3n)(2m – 3n)]
Put (2m + 3n) = a and (2m – 3n) = b. We get
Volume = a3 + b3 + 3(a + b) ab
Or, Volume = a3 + b3 + 3ab (a + b)
Or, Volume = (a + b)3 …(a)
[∵, by algebraic identity, we know that
(a + b)3 = a3 + b3 + 3ab (a + b)]
Again,
Put a = (2m + 3n) and b = (2m – 3n) in equation (a). We get
Volume = ((2m + 3n) + (2m – 3n))3
⇒ Volume = (2m + 3n + 2m – 3n)3
⇒ Volume = (4m)3
We know, volume of cube is given by
Volume = (length)3
Or (length)3 = Volume
⇒ length = (Volume)1/3
⇒ length = 4m
Thus, length of cube is 4m unit.
(vii). Given is,
Volume of cube (cubic unit) = (a + b)3 – (a – b)3 – 6b(a2 – b2)
We need to find the length of cube.
Volume of cube can be written as,
Volume = (a + b)3 – (a – b)3 – [3 × 2b × (a2 – b2)]
[∵, 6b = 3 × 2b]
⇒ Volume = (a + b)3 – (a – b)3 – [3 × 2b × (a + b)(a – b)]
[∵, By identity, (a2 – b2) = (a + b)(a – b)]
⇒ Volume = (a + b)3 – (a – b)3 – [3 × (b + b) × (a + b)(a – b)]
[∵, 2b = b + b]
⇒ Volume = (a + b)3 – (a – b)3 – [3 × (b + b + a – a) × (a + b)(a – b)]
[∵, a – a = 0 and it won’t affect the equation]
⇒ Volume = (a + b)3 – (a – b)3 – [3 × ((a + b) + (-a + b)) × (a + b)(a – b)]
[∵, (b + b + a – a) = ((a + b) + (-a + b)); we have just rearranged]
⇒ Volume = (a + b)3 – (a – b)3 – [3 × ((a + b) – (a – b)) × (a + b)(a – b)]
[∵, (-a + b) = -(a – b)]
⇒ Volume = (a + b)3 – (a – b)3 – 3((a + b) – (a – b))(a + b)(a – b)
Put (a + b) = x and (a – b) = y. We get
Volume = x3 – y3 – 3(x – y)xy
Or, Volume = x3 – y3 – 3xy(x – y)
Or, Volume = (x – y)3 …(A)
[∵, by algebraic identity, we know that
(x – y)3 = x3 – y3 – 3xy(x – y)]
Again,
Put x = (a + b) and y = (a – b) in equation (A). We get
Volume = ((a + b) – (a – b))3
⇒ Volume = (a + b – a + b)3
⇒ Volume = (2b)3
We know, volume of cube is given by
Volume = (length)3
Or (length)3 = Volume
⇒ length = (Volume)1/3
⇒ length = 2b
Thus, length of cube is 2b unit.
(viii). Given is,
Length of cube = 2x – 3y – 4z
Then,
Volume of cube = (2x – 3y – 4z)3
Recall the algebraic identity,
(a + b + c)3 = a3 + b3 + c3 + 3(a + b + c)(ab + bc + ca) – 3abc
Put a = 2x, b = -3y and c = -4z. We get
(2x – 3y – 4z)3 = (2x)3 + (-3y)3 + (-4z)3 + 3(2x – 3y – 4z)((2x)(-3y) + (-3y)(-4z) + (-4z)(2x)) – 3(2x)(-3y)(-4z)
⇒ (2x – 3y – 4z)3 = 8x3 – 27y3 – 64z3 + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz
Thus, volume of cube is [8x3 – 27y3 – 64z3 + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz] cubic unit.
(ix). Given is,
Volume of cube (cubic unit) = x6 – 15x4 + 75x2 – 125
We need to find length of cube.
Volume of cube can be written as,
[∵, x6 = (x2)3, 15 = 3 × 5, 75 = 3 × 25 and 125 = (5)3]
Volume = (x2)3 – (3 × 5x4) + (3 × 25x2) – (5)3
⇒ Volume = (x2)3 – (3 × 5 × (x2)2) + (3 × (5)2 × x2) – (5)3
[∵, x4 = (x2)2 and 25 = (5)2]
⇒ Volume = (x2)3 + (3 × -5 × (x2)2) + (3 × (-5)2 × x2) + (-5)3 …(A)
[∵, (5)2 = (-5)2 and -(5)3 = (-5)3]
Put x2 = a and (-5) = b in equation (A). We get
Volume = a3 + (3 × b × a2) + (3 × b2 × a) + b3
Or, Volume = a3 + 3ba2 + 3b2a + b3
Or, Volume = (a + b)3
[∵, by algebraic identity,
(a + b)3 = a3 + b3 + 3a2b + 3ab2]
Put a = x2 and b = (-5). We get
Volume = (x2 – 5)3
We know, volume of cube is given by
Volume = (length)3
Or (length)3 = Volume
⇒ length = (Volume)1/3
⇒ length = x2 – 5
Thus, length of cube is (x2 – 5) unit.
(x). Given is,
Volume of cube (cubic unit) = 1000 + 30x(10 + x) + x3
We need to find length of cube.
Volume of cube can be written as,
Volume = (10)3 + 30x(10 + x) + x3
[∵, 1000 = (10)3]
⇒ Volume = (10)3 + [3 × 10x × (10 + x)] + x3 …(A)
[∵, 30 = 3 × 10]
Put 10 = a and x = b in equation (A). We get
Volume = a3 + [3 × ab × (a + b)] + b3
Or, Volume = a3 + b3 + 3ab(a + b)
Or, Volume = (a + b)3
[∵, by algebraic identity,
(a + b)3 = a3 + b3 + 3ab(a + b)]
Put a = 10 and b = x. We get
Volume = (10 + x)3
We know, volume of cube is given by
Volume = (length)3
Or (length)3 = Volume
⇒ length = (Volume)1/3
⇒ length = 10 + x
Thus, length of cube is (10 + x) unit.
Question 2.
Let us solve the questions given below with the help of identities from I to IV.
a. Let’s write the value of if
b. Let’s try to prove if
c. Let’s write the value of by calculation if and
d. Let’s try to write the value of if
e. Let’s write the value of by calculation if
f. Let’s write the value of if
g. Let’s try to prove if
h. Let’s try to prove if
i. Let’s write the value of if
j. Let’s write the value of by calculation if
k. Let’s write the value of by calculation if and
Answer:
Let us solve.
(a). Given that, x – y = 2.
We need to find the value of x3 – y3 – 6xy.
By algebraic identity,
(a – b)3 = a3 – b3 – 3a2b + 3ab2
Or, (a – b)3 = a3 – b3 – 3ab(a – b)
Or, a3 – b3 = (a – b)3 + 3ab(a – b)
Put a = x and b = y. We get
x3 – y3 = (x – y)3 + 3xy(x – y) …(i)
Take x3 – y3 – 6xy.
x3 – y3 – 6xy = (x3 – y3) – 6xy
⇒ x3 – y3 – 6xy = ((x – y)3 + 3xy(x – y)) – 6xy [from (i)]
⇒ x3 – y3 – 6xy = ((2)3 + 3xy(2)) – 6xy [∵, Given that, (x – y) = 2]
⇒ x3 – y3 – 6xy = (8 + 6xy) – 6xy [∵, 23 = 8]
⇒ x3 – y3 – 6xy = 8 + 6xy – 6xy
⇒ x3 – y3 – 6xy = 8 + 0 [∵, 6xy – 6xy = 0]
⇒ x3 – y3 – 6xy = 8
Thus, x3 – y3 – 6xy = 8.
(b). Given:
To prove:
Proof: We know the algebraic identity,
(a + b)3 = a3 + b3 + 3a2b + 3ab2
Or, (a + b)3 = a3 + b3 + 3ab(a + b)
Or, a3 + b3 = (a + b)3 – 3ab(a + b) …(i)
Take a3 + b3 – ab.
a3 + b3 – ab = (a3 + b3) – ab
⇒ a3 + b3 – ab = ((a + b)3 – 3ab(a + b)) – ab
[∵, we know that, ]
Thus, .
Hence proved.
(c). Given that, x + y = 2 and
Let us solve this further.
We have,
⇒ x + y = 2xy
⇒ 2xy = x + y
⇒ 2xy = 2 [∵, we know that, (x + y) = 2]
⇒ xy = 1
Now, we have
x + y = 2 …(i)
xy = 1 …(ii)
We need to find the value of x3 + y3.
By algebraic identity,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
Or, (x + y)3 = x3 + y3 + 3xy(x + y)
Or, x3 + y3 = (x + y)3 – 3xy(x + y)
Substituting value of (x + y) and xy from equation (i) and (ii) respectively, we get
x3 + y3 = (2)3 – 3(1)(2)
⇒ x3 + y3 = 8 – 6
⇒ x3 + y3 = 2
Thus, x3 + y3 = 2.
(d). Given that,
…(i)
We need to find the value of
Further simplifying (i), we get
x2 – 1 = 2x
Now, take cube on both sides. We get
(x2 – 1)3 = (2x)3
[∵, we have the identity,
(a – b)3 = a3 – b3 – 3a2b + 3ab2]
⇒ (x2)3 – 13 – 3(x2)2 + 3x2 = (2x)3
⇒ x6 – 1 – 3x4 + 3x2 = 8x3
Dividing both sides by x3,
Thus, .
(e). Given that,
We need to find the value of .
Let us take .
Take cube on both sides of this equation. We get
[∵, from the identity, (a + b)3 = a3 + b3 + 3a2b + 3ab2]
Thus, .
(f). Given that,
x = y + z
We need to find the value of x3 – y3 – z3 – 3xyz.
Take x = y + z.
We can write as,
x – y – z = 0 …(i)
Now, take cube on both sides of the above equation. We get,
(x – y – z)3 = 0
Since, by the identity
(x – y – z)3 = x3 – y3 – z3 + 3(x – y – z)(-xy + yz – zx) – 3xyz
Rearranging it,
x3 – y3 – z3 – 3xyz = (x – y – z)3 – 3(x – y – z)(-xy + yz – zx)
⇒ x3 – y3 – z3 – 3xyz = (0)3 – 3(0)(-xy + yz – zx) [from equation (i)]
⇒ x3 – y3 – z3 – 3xyz = 0 – 0
⇒ x3 – y3 – z3 – 3xyz = 0
Thus, x3 – y3 – z3 – 3xyz = 0.
(g). Given: xy(x + y) = m
To prove:
Proof: Take xy(x + y) = m.
Rearranging it,
Taking cube on both sides, we get
By identity, we have
(x + y)3 = x3 + y3 + 3x2y + 3xy2
So,
[∵, given that, xy(x + y) = m]
Hence, proved.
(h). Given:
To prove:
Proof: We have,
Putting and .
⇒ p + q = 4pq …(i)
We have,
Multiplying p and q, we get
…(ii)
Now, take .
[∵, (p + q)3 = p3 + q3 + 3p2q + 3pq2
Or, p3 + q3 = (p + q)3 – 3p2q – 3pq2
Or, p3 + q3 = (p + q)3 – 3pq(p + q)]
…(iii)
Substituting the value of pq from equation (ii) in equation (iii), we get
Hence, proved.
(i). Given that,
We need to find the value of .
We have,
…(i)
Taking cube on both sides of the equation, we get
Now, adding 2 on both sides of this equation, we get
Thus, .
(j). Given that,
a3 + b3 + c3 = 3abc
where a ≠ b ≠ c.
We need to find the value of a + b + c.
We have,
a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 – 3abc = 0
We know that,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
⇒ 0 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
⇒ (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = 0
This means,
Either (a + b + c) = 0
Or (a2 + b2 + c2 – ab – bc – ca) = 0
If a2 + b2 + c2 – ab – bc – ca = 0
Multiplying by 2 on both sides, we get
2(a2 + b2 + c2 – ab – bc – ca) = 2 × 0
⇒ 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0
⇒ a2 + a2 + b2 + b2 + c2 + c2 – 2ab – 2bc – 2ca = 0
⇒ (a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 + 2ca) = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
[∵, by identity, we know (x – y)2 = x2 + y2 – 2xy]
This means,
(a – b)2 = 0 ⇒ a = b
(b – c)2 = 0 ⇒ b = c
(c – a)2 = 0 ⇒ c = a
But, a ≠ b ≠ c.
⇒ (a2 + b2 + c2 – ab – bc – ca) ≠ 0
This clearly means,
a + b + c = 0
Thus, a + b + c = 0.
(k). Given that,
m + n = 5 …(i)
mn = 6 …(ii)
We need to find the value of (mét vuông + n2)(m3 + n3).
By identity, we know
(m + n)3 = m3 + n3 + 3mn(m + n)
Or, m3 + n3 = (m + n)3 – 3mn(m + n) …(iii)
Also, by identity,
(m + n)2 = mét vuông + n2 + 2mn
Or, mét vuông + n2 = (m + n)2 – 2mn …(iv)
Take (mét vuông + n2)(m3 + n3).
So,
(mét vuông + n2)(m3 + n3) = ((m + n)2 – 2mn)((m + n)3 – 3mn(m + n))
Substituting values of (m + n) and mn from equation (i) and (ii) respectively in the above equation, we get
(mét vuông + n2)(m3 + n3) = ((5)2 – 2(6))((5)3 – 3(6)(5))
⇒ (mét vuông + n2)(m3 + n3) = (25 – 12)(125 – 90)
⇒ (mét vuông + n2)(m3 + n3) = 13 × 35
⇒ (mét vuông + n2)(m3 + n3) = 455
Thus, (mét vuông + n2)(m3 + n3) = 455.
Lets Work Out 5.3
Question 1.
Let us write in the blank box:
Answer:
(i)
Expression 1: x + 9
Expression 2: x2 – 9x + 81
Product
= (x + 9)(x2 – 9x + 81)
= (x + 9)(x2 – 9x + 92)
= x3 + 93 [∵ (a + b)(a2 – ab + b2) = a3 + b3]
= x3 + 729
(ii)
Expression 1: 2a – 1
Product = 8a3 – 1
= (2a)3 - 13
= (2a – 1)[(2a)2 + (2a)(1) + (1)2]
[∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= (2a – 1)(4a2 + 2a + 1)
Hence, Expression 2 is (4a2 + 2a + 1)
(iii)
Expression 1: 3 – 5c
Product = 27 – 125c3
= (3)3 – (5c)3
= (3 – 5c)(32 + 3(5c) + (5c)2) [∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= (3 – 5c)(9 + 15c + 25c2)
Hence, Expression 2 is (9 + 15c + 25c2)
(iv)
Expression 1: (a + b + c)
Expression 2: (a + b)2 – (a + b)c + c2
Product
= (a + b + c)[(a + b)2 – (a + b)c + c2]
= [(a + b) + c][(a + b)2 – (a + b)c + c2]
[Using, a3 + b3 = (a + b)(a2 – ab + b2)]
If a = a + b, b = c
= (a + b)3 + c3
(v)
Expression 1: 3x
Expression 2: (2x – 1)2 – (2x – 1) (x + 1) + (x + 1)2
Product
= 3x [(2x – 1)2 – (2x – 1)(x + 1) + (x + 1)2]
As 3x can be written as 2x-1+x+1
= [2x-1+x+1] [(2x – 1)2 – (2x – 1) (x + 1) + (x + 1)2]
[Using, a3 + b3 = (a + b) (a2 – ab + b2)
Here a = 2x-1, b = x+1
= (2x – 1)3 + (x + 1)3
As (a-b)3 =a3 -3a2b+3ab2 – b3
(a + b)3=a3+3a2b+3ab2+b3
= (2x)3 – 1 – 3(2x)2(1) + 3(2x)(1)2 + x3 + 1 + 3x2 + 3x
= 8x3 – 1 – 12x2 + 6x + x3 + 1 + 3x2 + 3x
= 9x3 – 9x2 + 9x
(vi)
Expression 1:
Expression 2:
Product
[using, (a + b)3 = (a + b)(a2 – ab + b2)]
(vii)
Expression 1: 4a – 5b
Expression 2: 16a2 + 20ab + 25b2
Product
= (4a – 5b)(16a2 + 20ab + 25b2)
= (4a – 5b)((4a)2 + (4a)(5b) + (5b)2)
[Using, a3 – b3 = (a – b)(a2 + ab + b2)]
= (4a)3 – (5b)3
= 64a3 – 125b3
(viii)
Expression 2: a2b2 + abcd + c2d2
Product
= a3b3 – c3d3
= (ab)3 – (cd)3
= (ab – cd)[(ab)2 + (ab)(cd) + (cd)2]
[Using, a3 – b3 = (a – b)(a2 + ab + b2)]
= (ab – cd)(a2b2 + abcd + c2d2]
Hence, Expression 1 is (ab – cd)
(ix)
Expression 1: 1 – 4y
Product = 1 – 64y3
= 13 – (4y)3
= (1 – 4y)(12 + 1(4y) + (4y)2)
[∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= (1 – 4y)(1 + 4y + 16y2)
Hence, Expression 2 is (1 + 4y + 16y2)
(x)
Expression 1: (2p + 1)
Product = 8(p – 3)3 + 343
= (2(p – 3))3 + 73
[Using, a3 – b3 = (a – b)(a2 + ab + b2)]
= [2(p – 3) + 7][(2(p – 3))2 + 2(p – 3)(7) + 72]
= (2p – 6 + 7)[4(p – 3)2 + 14p – 42 + 49]
= (2p + 1)(4(p2 – 6p + 9) + 14p + 7)
= (2p + 1)(4p2 – 24p + 36 + 14p + 7)
= (2p + 1)(4p2 – 10p + 43)
Hence, Expression 2 is (4p2 – 10p + 43)
(xi)
Expression 1 : m – p
Expression 2: (m + n)2 + (m + n)(n + p) + (n + p)2
Product
= (m – p)[(m + n)2 + (m + n)(n + p) + (n + p)2]
= [m + n – (n + p)]((m + n)2 + (m + n)(n + p) + (n + p)2]
[Using, a3 – b3 = (a – b)(a2 + ab + b2)]
= (m + n)3 – (n + p)3
= (m3 + n3 + 3m2n + 3mn2) – (n3 + p3 + 3n2p + 3np2)
= m3 + n3 + 3m2n + 3mn2 – n3 – p3 – 3n2p – 3np2
= m3 – p3 + 3m2n + 3mn2 – 3n2p – 3np2
(xii)
Expression 1: (3a - 2b)2 + (3a – 2b) × (2a – 3b) + (2a – 3b)2
Expression 2: (a + b)
Product
= (a + b)[(3a - 2b)2 + (3a – 2b) × (2a – 3b) + (2a – 3b)2]
= (3a – 2b – (2a – 3b))[(3a - 2b)2 + (3a – 2b) × (2a – 3b) + (2a – 3b)2]
= (3a – 2b)3 - (2a – 3b)3
[Now,
(a – b)3 = a3 – b3 – 3a2b + 3ab2]
= [(3a)3 – (2b)3 – 3(3a)2(2b) + 3(3a)(2b)2] – [(2a)3 – (3b)3 – 3(2a)2(3b) + 3(2a)(3b)2]
= [27a3 – 8b3 – 54a2b + 36ab2] – [8a3 – 27b3 – 36a2b + 54ab2]
= 27a3 – 8b3 – 54a2b + 36ab2 – 8a3 + 27b3 + 36a2b – 54ab2
= 19a3 + 19b3 – 18a2b – 18b2
Question 2.
Let us simplify using formula:
i.
ii.
iii.
iv.
v.
Answer:
i.(a + b)(a – b)(a2 + ab + b2)(a2 – ab + b2)
= (a + b)(a2 – ab + b2)(a – b)(a2 + ab + b2)
Using x3 – y3 = (x – y)(x2 + xy + y2)
x3 + y3 = (x + y)(x2 – xy + y2)
= (a3 + b3)(a3 – b3)
Now, using (a + b)(a – b) = a2 – b2
= a6 – b6
ii. (a – 2b)(a2 + 2ab + 4b2)(a3 + 8b3)
= (a – 2b)(a2 + a(2b) + (2b)2)(a3 + (2b)3)
Using x3 – y3 = (x – y)(x2 + xy + y2)
x3 + y3 = (x + y)(x2 – xy + y2)
= (a3 – (2b)3)(a3 + (2b)3)
Now, using (a + b)(a – b) = a2 – b2
= a6 – (2b)6
= a6 – 64b6
iii. (4a2 – 9)(4a2 – 6a + 9)(4a2 + 6a + 9)
= [(2a)2 – 32](4a2 – 6a + 9)(4a2 + 6a + 9)
Using x2 – y2 = (x – y)(x + y)
= (2a – 3)(2a + 3)(4a2 – 6a + 9)(4a2 + 6a + 9)
= (2a – 3)(4a2 + 6a + 9)(2a + 3)(4a2 – 6a + 9)
Using x3 – y3 = (x – y)(x2 + xy + y2)
x3 + y3 = (x + y)(x2 – xy + y2)
= [(2a)3 – 33][(2a)3 + 33]
= (2a)6 - 36
= 64a6 – 729
iv. (x – y)(x2 + xy + y2) + (y – z)(y2 + yz + z2) + (z – x)(z2 + zx + x2)
Using a3 – b3 = (a – b)(a2 + ab + b2)
= x3 – y3 + y3 – z3 + z3 – x3
= 0
v. (x + 1)(x2 - x + 1) + (2x – 1)(4x2 + 2x + 1) – (x – 1)(x2 + x + 1)
= (x + 1)(x2 - x + 1) + (2x – 1)((2x)2 + 2x + 1) – (x – 1)(x2 + x + 1)
Using x3 – y3 = (x – y)(x2 + xy + y2)
x3 + y3 = (x + y)(x2 – xy + y2)
= x3 + 1 + (2x)3 – 1 - (x3 – 1)
= x3 + 1 + 8x3 – 1 – x3 + 1
= 8x3 + 1
Question 3.
Let’s find the value of by calculation if
or,
or,
or, [by transposition]
[With the help of the identity No. VI]
Answer:
Given,
Multiplying both side by ‘x’
⇒ x2 + 1 = -x
⇒ x2 + x + 1 = 0
Now,
x3 – 1
Using x3 – y3 = (x – y)(x2 + xy + y2)
= (x – 1)(x2 + x + 1)
= (x – 1) × 0
= 0
Question 4.
Let’s find the value of by calculation if
Answer:
Given,
Multiplying both side by ‘a’
⇒ a2 + 9 = 3a
⇒ a2 – 3a + 9 = 0
Now,
a3 + 27
= a3 + 33
Using x3 + y3 = (x + y) (x2 – xy + y2)
= (a + 3) (a2 – 3a + 32)
= (a + 3) (a2 – 3a + 9)
= (a + 3) × 0
= 0
Question 5.
Let’s find the value of by calculation if
Answer:
Given,
Multiplying both side by ‘ab’
⇒ a2 + b2 = ab
⇒ a2 – ab + b2 = 0
Now,
a3 + b3
= (a + b)(a2 – ab + b2)
= (a + b) × 0
= 0
Question 6.
Let’s resolve into factors the following algebraic expressions:
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
xiii.
xiv.
xv.
xvi.
xvii.
Answer:
(i) 1000a3+27b6
= (10a)3 + (3b2)6
Using x3 + y3 = (x = y)(x2 - xy + y2)
= (10a + 3b2)[(10a)2 - (10a)(3b2) + (3b2)2]
= (10a + 3b2)(100a2 - 30ab2 + 9b4)
(ii) 1-216 z3
= 1 – (6z)3
Using x3 – y3 = (x – y)(x2 + xy + y2)
= (1 – 6z)(12 + 1(6z) + (6z)2)
= (1 – 6z)(1 + 6z + 36z2)
(iii) m4 - m
= m(m3 – 1)
= m(m3 – 13)
Using x3 – y3 = (x – y)(x2 + xy + y2)
= m(m – 1)(mét vuông + m + 1)
(iv) 192a3+3
= 3(64a3 + 1)
= 3[(4a)3 + 13]
Using x3 + y3 = (x + y)(x2 – xy + y2)
= 3(4a + 1)[(4a)2 – 4a + 1]
= 3(4a + 1)(16a2 – 4a + 1)
(v) 16a4x3 + 54ay3
= 2a(8a3x3 + 27y3)
= 2a[(2ax)3 + (3y)3]
Using x3 + y3 = (x + y)(x2 – xy + y2)
= 2a(2ax + 3y)[(2ax)2 – 2ax(3y) + (3y)2]
= 2a(2ax + 3y)(4a2x2 – 6axy + 9y2)
(vi) 729a3 b3 c3-125
= (9abc)3 - 53
= (9abc – 5)[(9abc)2 + 9abc(5) + 52]
= (9abc – 5)(81a2b2c2 + 45abc + 25)
(vii)
Using x3 - y3 = (x - y)(x2 + xy + y2)
(viii)
Using x3 – y3 = (x – y)(x2 + xy + y2)
(ix)x3 + 3x2y + 3xy2 + 2y3
= x3 + y3 + 3x2y + 3xy2 + y3
= (x + y)3 + y3
[∵ a3 + b3 + 3a2b + 3ab2 = (a + b)3]
= (x + y + y)[(x + y)2 - (x + y)y + y2]
[Using x3 – y3 = (x + y)(x2 - xy + y2)]
= (x + 2y)(x2 + y2 - xy - y2 + y2)
= (x + 2y)(x2 – xy + y2)
(x) 1 + 9x + 27x2 + 28x3
= 27x3 + 1 + 27x2 + 9x + x3
= (3x)3 + 13 + 3(3x)3(1) + 3(3x)(1)2
= (3x + 1)3 + x3
[∵ a3 + b3 + 3a2b + 3ab2 = (a + b)3]
Now, Using x3 + y3 = (x + y)(x2 - xy + y2)
= (3x + 1 + x)[(3x + 1)2 - (3x + 1)x + x2]
= (4x + 1)(9x2 + 6x + 1 - 3x2 - x + x2)
= (4x + 1)(7x2 + 5x + 1)
(xi) x3 – 9y3 – 3xy(x-y)
x3 – y3 – 3xy (x – y) – 8y3
[∵ x3 – y3 = x3 – y3 – 3xy (x – y)]
= (x – y)3 – (2y)3
= (x – y – (2y)) [(x – y)2 + (x – y) (2y) + (2y)2]
[Using x3 + y3 = (x + y) (x2 – xy + y2)]
= (x -3 y) (x2 + y2 – 2xy + 2xy – 2y2 + 4y2)
= (x - 3y) (x2 + 3y2)
(xii)8 – a3 + 3a2b – 3ab2 + b3
= b3 – a3 – 3ab2 + 3a2b + 8
Now, As (x + y)3 = x3 + y3 + 3x2y + 3xy2
= (b – a)3 + 23
Using x3 + y3 = (x + y) (x2 - xy + y2)
= (b – a + 2) [(b – a)2 - (b – a)2 + 22]
= (b – a + 2) (b2 + a2 – 2ab - 2b + 2a + 4)
(xiii) x6+3x4 b2+3x2 b4+b6+a3b3
= (x2)3 + (b2)3 + 3(x2)2(b2) + 3(x2)(b2)2 + a3b3
As (x + y)3 = x3 + y3 + 3x2y + 3xy2
= (x2 + b2)3 + (ab)3
Using x3 + y3 = (x + y)(x2 – xy + y2)
= (x2 + b2 + ab)[(x2 + b2)2 - (x2 + b2)ab + a2b2]
(xiv) x6 + 27
= (x2)3 + (3)3
Using x3 + y3 = (x + y)(x2 – xy + y2)
= (x2 + 3)[(x2)2 - 3x2 + 32)
= (x2 + 3)(x4 - 3x2 + 9)
(xv)x6 – y6
= (x3)2 – (y3)2
Using (a2 – b2) = (a – b)(a + b)
= (x3 – y3)(x3 + y3)
Using x3 – y3 = (x – y)(x2 + xy + y2) and
x3 + y3 = (x + y)(x2 – xy + y2)
= (x – y)(x2 + xy + y2)(x + y)(x2 – xy + y2)
(xvi) x12 – y12
= (x6)2 – (y6)2
Using (a2 – b2) = (a – b)(a + b)
= (x6 – y6)(x6 + y6)
= [(x3)2 – (y3)2 �][(x2)3 + (y2)3]
Using (a2 – b2) = (a – b)(a + b)
= (x3 – y3)(x3 + y3)[(x2)3 + (y2)3]
Using x3 – y3 = (x – y)(x2 + xy + y2) and
x3 + y3 = (x + y)(x2 – xy + y2)
= (x – y)(x2 + xy + y2)(x + y)(x2 – xy + y2)(x2 + y2)(x4 - x2y2 + x6y6)
(xvii) m3 -n3-m(mét vuông-n2 )+n(m-n)2
Using x3 – y3 = (x – y)(x2 + xy + y2) and
Using (a2 – b2) = (a – b)(a + b)
= (m – n)(mét vuông + mn + n2) – m(m – n)(m + n) + n(m – n)2
Taking (m – n) as common, we get
= (m – n)[m2 + mn + n2 – m(m + n) + n(m – n)]
= (m – n)(mét vuông + mn + n2 – mét vuông – mn + mn – n2)
= (m – n)mn
What should be multiplied by 1225 to make a perfect cube?
Answer. 1225 × 35 = 42875 which is a perfect cube.What is the smallest number that should be multiplied to make it a perfect cube?
Suppose if we have 13 in the form of a cube, then it will be a perfect cube. To make the 1352 as a perfect we have to multiply the number by 13. Therefore 13 is the smallest number which should be multiplied to 1352 to get a perfect cube. So, the correct answer is “ 13”.Is 7803 a perfect cube?
It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube.Is 10976 a perfect cube?
This is Expert Verified Answer So we need to multiply another 2 to the factorization to make 10976 a perfect cube. Tải thêm tài liệu liên quan đến nội dung bài viết What is the smallest number that 1225 must be multiplied by to make it a perfect cube?