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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) \(100\)

 Video Solution
Cubes And Cube Roots
Ex 7.1 | Question 2

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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