Question

Find the cube root of the following rational number \[\frac{- 39304}{- 42875}\] .

Answer 1

Let us consider the following rational number:

\[\frac{- 39304}{- 42875}\]

Now

\[\sqrt[3]{\frac{- 39304}{- 42875}}\]

\[= \frac{\sqrt[3]{- 39304}}{\sqrt[3]{- 42875}}\]   ( \[\because\] \[\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}\] ) 

\[= \frac{- \sqrt[3]{39304}}{- \sqrt[3]{42875}}\] ( ∵ \[\sqrt[3]{- a} = - \sqrt[3]{a}\] Cube root by factors:

On factorising 39304 into prime factors, we get:

\[39304 = 2 \times 2 \times 2 \times 17 \times 17 \times 17\]

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

\[39304 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 17 \times 17 \times 17 \right\}\]

Now, taking one factor from each triple, we get:

 

\[\sqrt[3]{39304} = 2 \times 17 = 34\]

Also
On factorising 42875 into prime factors, we get:

\[42875 = 5 \times 5 \times 5 \times 7 \times 7 \times 7\]

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

 

\[42875 = \left\{ 5 \times 5 \times 5 \right\} \times \left\{ 7 \times 7 \times 7 \right\}\]

Now, taking one factor from each triple, we get:

\[\sqrt[3]{42875} = 5 \times 7 = 35\]

Now

\[\sqrt[3]{\frac{- 39304}{- 42875}}\]

\[= \frac{\sqrt[3]{- 39304}}{\sqrt[3]{- 42875}}\]

\[= \frac{- \sqrt[3]{39304}}{- \sqrt[3]{42875}}\]

\[= \frac{- 34}{- 35}\]

\[ = \frac{34}{35}\]