Ex.7.1 Q2 Cubes and Cube Roots - NCERT Maths Class 8

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Question

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) \(100\)

Text Solution

What is unknown?

To find the smallest number by which the given number must be multiplied 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, the smallest number can be identified.

Steps:

(i) 

\[\begin{align}243 &= \underline {3 \times 3 \times 3}  \times 3 \times 3\\243 &= {3^3} \times 3\end{align}\]

Here, one of the \(3'\rm{s}\) is  not a triplet.To make it as a triplet, we need to multiply by \(3\) 

In that case,

 \(\begin{align}243 \times 3 = \underline {3 \times 3 \times 3}  \times \underline {3 \times 3 \times 3}  = {3^3} \times {3^3} = {9^3} = 729\end{align}\) is a perfect cube.

Hence, the smallest natural number by which \(243\) should be multiplied to make a perfect cube is \(3\).

(ii)

\[\begin{align}256 &= \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times 2 \times 2\\256 &= {2^3} \times {2^3} \times {2^2}\end{align}\]

Here one of the \( 2’\rm{s}\) is not a triplet. To make it as a triplet, we need to multiply by \(2.\)

In that case,

 \(\begin{align}256 \times 2 = \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  = {2^3} \times {2^3} \times {2^3} = {8^3} = 512\end{align}\) is a perfect cube.

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

(iii) 

\[\begin{align}72 &= \underline {2 \times 2 \times 2}  \times 3 \times 3\\72 &= {2^3} \times {3^2}\end{align}\]

Here,on of the \(3\)'s is not a triplet. To make it as a triplet, we need to multiply by \(3.\)

In that case, 

\(\begin{align}72 \times 3 = \underline {2 \times 2 \times 2}  \times \underline {3 \times 3 \times 3}  = {2^3} \times {3^3} = {6^3} = 216\end{align}\) is a perfect cube.

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

(iv) 

\[\begin{align}675 &= 5 \times 5 \times \underline {3 \times 3 \times 3} \\675 &= {5^2} \times {3^3}\end{align}\]

Here, one of the \(5\)'s is not a triplet. To make it as a triplet, we need to multiply by \(5.\)

In that case,

\(\begin{align}675 = \underline {5 \times 5 \times 5}  \times \underline {3 \times 3 \times 3}  = {5^3} \times {3^3} = {15^3} = 3375\end{align}\) is a perfect cube.

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

(v)

\[\begin{align}100 &= 2 \times 2 \times 5 \times 5\\100 &= {2^2} \times {5^2}\end{align}\]

Here both the prime factors are not triplets. To make them triplets we need to multiply by one more  \(2\) and \(5.\)

In that case,

\(\begin{align}100 = \underline {2 \times 2 \times 2}  \times \underline {5 \times 5 \times 5}  = {2^3} \times {5^3} = {10^3} = 1000\end{align}\) is a perfect cube.

 Hence, the smallest natural number by which 100 should be multiplied to make a perfect cube is \(2 \times 5 = 10.\)

 

 

  
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