# Ex.7.1 Q3 Cubes and Cube Roots - NCERT Maths Class 8

## Question

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) \(704\)

## Text Solution

**What is unknown?**

To find the smallest number by which a given number must be divided to obtain a perfect cube.

**Reasoning:**

A number is a perfect cube only when each factor in the prime factorization is grouped in triples. Using this concept smallest number to be multiplied can be obtained.

(i)

\[\begin{align}81 &= \underline {3 \times 3 \times 3} \times 3\\81 &= {3^3} \times 3\end{align}\]

Here, the prime factor \(3\) is not present as triples.

Hence, we divide by \(81\) by \(3,\) so that the obtained number becomes a perfect cube.

Thus,

\(\begin{align}81 \div 3 = 27 = {3^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(81\) should be divided to make a perfect cube is \(3\).

(ii)

\[\begin{align}128& = \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \times 2\\128 &= {2^3} \times {2^3} \times 2\end{align}\]

Here,the prime factors \(2\) is not present as triples.

Hence, we divide \( 128\) by \(2,\) so that the obtained number becomes a perfect cube.

\(\begin{align}128 \div 2 = 64 = {2^3} \times {2^3} = {4^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(128\) should be divided to make a perfect cube is \(2\).

(iii)

\[\begin{align}135 &= 5 \times \underline {3 \times 3 \times 3} \\135 &= {5^1} \times {3^3}\end{align}\]

Here, the prime factors \(5\) is not present as triples.

Hence, we divide \(135\) by \(5,\) so that the obtained number becomes a perfect cube.

\(\begin{align}135 \div 5 = 27 = {3^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(135\) should be divided to make a perfect cube is \(5\).

(iv)

\[\begin{align}192 &= \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \times 3\\192 &= {2^3} \times {2^3} \times 3\end{align}\]

Here,the prime factors \(3\) is not present as triples.

Hence, we divide \(192\) by \(3,\) so that the obtained number becomes a perfect cube.

\(\begin{align}192 \div 3 = 64 = {2^3} \times {2^3} = {4^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(192\) should be divided to make a perfect cube is \(3\).

(v)

\[\begin{align}704 &= \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \times 11\\704 &= {2^3} \times {2^3} \times 11\end{align}\]

Here, the prime factors \(11\) is not present as triples.

Hence, we divide \(704\) by \(11,\) so that the obtained number becomes a perfect cube.

\(\begin{align}704 \div 11 = 64 = {2^3} \times {2^3} = {4^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(704\) should be divided to make a perfect cube is \(11\).