NCERT Solutions Class 8 Maths Chapter 7

Q.1 Which of the following numbers are not perfect cubes?

(i) 216 (ii) 128 (iii) 1000 (iv)100 (v) 46656

Ans

$\begin{array}{l}\text{(i)The prime factorisation of 216 is as follows:}\\ \begin{array}{cc}2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ \text{216}=\underset{¯}{\text{2}\times \text{2}\times \text{2}}\times \underset{¯}{\text{3}\times \text{3}\times \text{3}}\\ \text{Since all the factors appears in a group of three,so 216}\\ \text{is a perfect cube}.\end{array}$

$\begin{array}{l}\text{(ii)The prime factorisation of 128 is as follows:}\\ \begin{array}{cc}2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ 128=\underset{¯}{\text{2}\times \text{2}\times \text{2}}\times \underset{¯}{\text{2}\times \text{2}\times \text{2}}\times 2\\ \text{Since 2 is not appearing in a group of three,so 128}\\ \text{is not a perfect cube}.\\ \\ \text{(iii)The prime factorisation of 1000 is as follows:}\\ \begin{array}{cc}2& 1000\\ 2& 500\\ 2& 250\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 1000=\underset{¯}{\text{2}\times \text{2}\times \text{2}}\times \underset{¯}{\text{5}\times \text{5}\times \text{5}}\\ \text{Since 2 and 5 appears in a group of three,so 1000}\\ \text{is a perfect cube}.\end{array}$

$\begin{array}{l}\text{(iv)The prime factorisation of 100 is as follows:}\\ \begin{array}{cc}2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 100=2\times 2\times 5\times 5\\ \text{Since 2 and 5 are not appearing in a group of three,so}\\ \text{100 is not a perfect cube}.\\ \\ \text{(v)The prime factorisation of 46656 is as follows:}\\ \begin{array}{cc}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}46656=\underset{¯}{\text{2}\times \text{2}\times \text{2}}\times \underset{¯}{\text{2}\times \text{2}\times \text{2}}\times \underset{¯}{3\times 3\times 3}\times \underset{¯}{3\times 3\times 3}\\ \text{Since all the factors appears in a group of three,so}\\ \text{46656 is a perfect cube}.\end{array}$

Q.2 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100

Ans

(i)

$\begin{array}{l}\text{The prime factorisation of 243 is as follows:}\\ \begin{array}{cc}3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 243=\underset{¯}{3\times 3\times 3}\times 3\times 3\end{array}$

Here, two 3s are left which does not appear in a group of three. To make 243 a cube, one more 3 is required.
In that case, we have to multiply 243 by 3
i.e. 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729, which is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii)

$\begin{array}{l}\text{The prime factorisation of 256 is as follows:}\\ \begin{array}{cc}2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ 256=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times 2\times 2\end{array}$

Here, two 2s are left which does not appear in a group of three. To make 256 a cube, one more 2 is required.

If we multiply 256 by 2, we get

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512, which is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii)

$\begin{array}{l}\text{The prime factorisation of 72 is as follows:}\\ \begin{array}{cc}2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 72=\underset{¯}{2\times 2\times 2}\times 3\times 3\end{array}$

Here, two 3s are left which are not in a group of three. To make 72 a perfect cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216, which is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv)

$\begin{array}{l}\text{The prime factorisation of 675 is as follows:}\\ \begin{array}{cc}3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\text{675}=\underset{¯}{\text{3}\times \text{3}\times \text{3}}\times \text{5}\times \text{5}$

Here, two 5s are left which are not in a group of three. To make 675 a cube, one more 5 is required.

If we multiply 675 by 5, we get

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375, which is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v)

$\begin{array}{l}\text{The prime factorisation of 100 is as follows:}\\ \begin{array}{cc}2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ \text{1}00=\text{2}\times \text{2}\times \text{5}\times \text{5}\end{array}$

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000, which is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

Q.3 Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704

Ans

(i)

$\begin{array}{l}\text{The prime factorisation of 81 is as follows:}\\ \begin{array}{cc}3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

81 = 3 × 3 × 3 × 3

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Therefore, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii)

$\begin{array}{l}\text{The prime factorisation of 128 is as follows:}\\ \begin{array}{cc}2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, one 2 is left which is not in a group of three.

If we divide 128 by 2, then it will become a perfect cube.

Therefore, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii)

$\text{The prime factorisation of 135 is as follows:}$

$\begin{array}{cc}3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$

135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a group of three.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(iv)

$\text{The prime factorisation of 192 is as follows:}$

$\begin{array}{cc}2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a group of three.

If we divide 192 by 3, then it will become a perfect cube.

Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(v)

$\text{The prime factorisation of 704 is as follows:}$

$\begin{array}{cc}2& 704\\ 2& 352\\ 2& 176\\ 2& 88\\ 2& 44\\ 2& 22\\ 11& 11\\ & 1\end{array}$

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a group of three.

If we divide 704 by 11, then it will become a perfect cube.

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

Q.4 Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Ans

Volume of the cube of sides 5 cm,2 cm ,5 cm

=5 cm ×5 cm ×2 cm =5 ×5 ×2 cm³

Here, two 5s and one 2 are left which are not in a triplet. If we multiply this expression by 5 × 2 × 2 = 20, then it will become a perfect cube.

Thus, 5 × 5 × 2 × 5 × 2 × 2 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, 20 cuboids of 5 cm, 2 cm, and 5 cm are required to form a cube.

Q.5 Find the cube root of each of the following numbers by prime factorization method.

(i) 64 (ii) 512 (iii) 10648 (iv)27000 (v) 15625
(vi)13824 (vii) 110592 (viii) 46656 (ix) 175616
(x) 91125

Ans

$\begin{array}{l}\text{(i) The prime factorisation of 64 is as follows:}\\ \begin{array}{cc}2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$

$\begin{array}{l}64=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\\ \\ \therefore \sqrt[3]{64}=2\times 2=4\end{array}$

$\begin{array}{l}\text{(ii) The prime factorisation of 512 is as follows:}\\ \begin{array}{cc}2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$

$\begin{array}{l}512=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\\ \\ \therefore \sqrt[3]{512}=2\times 2\times 2=8\end{array}$

$\begin{array}{l}\text{(iii) The prime factorisation of 10648 is as follows:}\\ \begin{array}{cc}2& 10648\\ 2& 5324\\ 2& 2662\\ 11& 1331\\ 11& 121\\ 11& 11\\ & 1\end{array}\end{array}$

$\begin{array}{l}10648=\underset{¯}{2\times 2\times 2}\times \underset{¯}{11\times 11\times 11}\\ \\ \therefore \sqrt[3]{10648}=2\times 11=22\end{array}$

$\begin{array}{l}\text{(iv)The prime factorisation of 27000 is as follows:}\\ \begin{array}{cc}2& 27000\\ 2& 13500\\ 2& 6750\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\begin{array}{l}27000=\underset{¯}{2\times 2\times 2}\times \underset{¯}{3\times 3\times 3}\times \underset{¯}{5\times 5\times 5}\\ \\ \therefore \sqrt[3]{27000}=2\times 3\times 5=30\end{array}$

$\begin{array}{l}\text{(v) The prime factorisation of 15625 is as follows:}\\ \begin{array}{cc}5& 15625\\ 5& 3125\\ 5& 625\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\begin{array}{l}15625=\underset{¯}{5\times 5\times 5}\times \underset{¯}{5\times 5\times 5}\\ \\ \therefore \sqrt[3]{15625}=5\times 5=25\end{array}$

$\begin{array}{l}\text{(vi) The prime factorisation of 13824 is as follows:}\\ \begin{array}{cc}2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}13824=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{3\times 3\times 3}\\ \\ \therefore \sqrt[3]{13824}=2\times 2\times 2\times 3=24\end{array}$

$\begin{array}{l}\text{(vii) The prime factorisation of 110592 is as follows:}\\ \\ \begin{array}{cc}2& 110592\\ 2& 55296\\ 2& 27648\\ 2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}110592=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{3\times 3\times 3}\\ \\ \therefore \sqrt[3]{110592}=2\times 2\times 2\times 2\times 3=48\end{array}$

$\begin{array}{l}\text{(viii) The prime factorisation of 46656 is as follows:}\\ \\ \begin{array}{cc}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}46656=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{3\times 3\times 3}\times \underset{¯}{3\times 3\times 3}\\ \\ \therefore \sqrt[3]{46656}=2\times 2\times 3\times 3=36\end{array}$

$\begin{array}{l}\text{(ix) The prime factorisation of 175616 is as follows:}\\ \\ \begin{array}{cc}2& 175616\\ 2& 87808\\ 2& 43904\\ 2& 21952\\ 2& 10976\\ 2& 5488\\ 2& 2744\\ 2& 1372\\ 2& 686\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\end{array}$

$\begin{array}{l}175616=\underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{2\times 2\times 2}\times \underset{¯}{7\times 7\times 7}\\ \\ \therefore \sqrt[3]{175616}=2\times 2\times 2\times 7=56\end{array}$

$\begin{array}{l}\text{(x) The prime factorisation of 91125 is as follows:}\\ \\ \begin{array}{cc}3& 91125\\ 3& 30375\\ 3& 10125\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\begin{array}{l}91125=\underset{¯}{3\times 3\times 3}\times \underset{¯}{3\times 3\times 3}\times \underset{¯}{5\times 5\times 5}\\ \\ \therefore \sqrt[3]{91125}=3\times 3\times 5=45\end{array}$

Q.6 State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv)There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi)The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

Ans

1. False.

Reason: The cube of any odd number is an odd number because when we find the cube of any odd number then we are multiplying its unit’s digit three times and the unit’s digit of any odd number is also an odd number. Therefore, the product will again be an odd number.

For example: The cube of 9 (i.e., an odd number) is 729, which is again an odd number.

2. True.

Reason: A perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3.

For example: The cube of 10 is 1000 and there are 3 zeroes at the end of it.

The cube of 100 is 1000000 and there are 6 zeroes at the end of it.

3. False.

Reason: It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.

For example: the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.

4. False.

Reason: The cubes of all the numbers having their unit’s digit as 2 will end with 8.

For example: The cube of 22 is 10648 and the cube of 32 is 32768.

5. False.

Reason: The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.

6. False.

Reason: The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(vii)True

Reason: As the cube of 1 and 2 are 1 and 8 respectively.

Q.7 You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

Ans

Firstly, we will make groups of three digits starting from the rightmost digit of the number 1331.

There are 2 groups, 1 and 331, in it.

Consider the first group 331.

The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.

Now we will take the second group i.e., 1.

The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the one’s place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.

$\text{Hence,}\sqrt[3]{1331}=11$ $\sqrt[3]{4913}$

First make groups of three digits starting from the rightmost digit of 4913.

The groups are 4 and 913.

Consider the first group 913.

The number 913 ends with 3. We know that if a perfect cube number ends with 3, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.

Now, take the second group i.e., 4.

We know that, 1³ = 1 and 2³ = 8

Also, 1 < 4 < 8

Therefore, we take the one’s place, of the smaller number 1 as the ten’s place of the required cube root.

$\text{Hence,}\sqrt[3]{4913}=17$

Now,

$\sqrt[3]{12167}$

First make groups of three digits starting from the rightmost digit of 12167.

The groups are 12 and 167.

Consider the first group 167.

167 end with 7. We know that if a perfect cube number ends with 7, then its cube root will have its unit’s digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.

Take the second group i.e., 12.

We know that, 2³ = 8 and 3³ = 27

Also, 8 < 12 < 27

2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.

$\text{Hence,}\sqrt[3]{12167}=23$ $\sqrt[3]{32768}$

We will make groups of three digits starting from the rightmost digit of the number 32768.

The groups are 32 and 768.

Consider the first group 768.

768 end with 8. We know that if a perfect cube number ends with 8, then its cube root will have its unit’s digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.

Taking the other group i.e., 32,

We know that, 3³ = 27 and 4³ = 64

Also, 27 < 32 < 64

3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.

$\text{Hence,}\sqrt[3]{32768}=32$