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Find the Cube Root of the Following Natural Number 134217728 . - Mathematics

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Question

Find the cube root of the following natural number 134217728 .

Sum

Solution

Cube root by factors:
On factorising 134217728 into prime factors, we get:

\[134217728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

On grouping the factors in triples of equal factors, we get:

\[134217728 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\}\]
Now, taking one factor from each triple, we get:
\[\sqrt[3]{134217728} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512\]
 
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Chapter 4: Cubes and Cube Roots - Exercise 4.3 [Page 22]

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RD Sharma Mathematics [English] Class 8
Chapter 4 Cubes and Cube Roots
Exercise 4.3 | Q 4.07 | Page 22

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