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Ex.7.1 Q1 Cubes and Cube Roots - NCERT Maths Class 8

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Question

Which of the following numbers are not perfect cubes?

(i) \(216\)         

(ii) \(128\)         

(iii) \(1000\)         

(iv) \(100\)         

(v) \(46656\)

 Video Solution
Cubes And Cube Roots
Ex 7.1 | Question 1

Text Solution

What is unknown?

To find the numbers which are not perfect cubes.

Reasoning:

A number is a prefect cube only when each factor in the prime factorization is grouped in triples.

Steps: 

(i)

\[\begin{align}216 &= {2 \times 2 \times 2} \times {3 \times 3 \times 3} \\&= {2^3} \times {3^3}\\& ={{(2\times 3)}^{3}} \\& ={{6}^{3}} \\\end{align}\]

\(\therefore 216\) is a perfect cube

 (ii)

\[\begin{align}128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \end{align}\]

\(128 = {2^3} \times {2^3} \times 2\) one of the \(2\) is not grouped in triples

\(\therefore \;128\) is not a perfect cube

 (iii)   

\[\begin{align}1000 &= 2 \times 2 \times 2 \times 5 \times 5 \times 5\\
1000 &= {2^3} \times {5^3}\end{align}\]

\(\therefore\;1000\) is a perfect cube

(iv) 

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

Both \(2\) and \(5\) are not grouped in triples

\(\therefore \;100\) is not a perfect cube.

(iv) 

\(\begin{align}46656 &=  \left(\begin{array} \ \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times \\ \underline {3 \times 3 \times 3}  \times \underline {3 \times 3 \times 3}\end{array} \right) \\ &=  {2^3} \times {2^3} \times {3^3} \times {3^3}
\end{align}\)

\(\therefore\;46656\) is a perfect cube

  
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