Question

Find the cube root of each of the following integers :

$\left(v\right)-32678$

Answer 1

Step: Find the cube root of $-32768$

$\begin{array}{cc}\text{2}& \text{32768}\\ \text{2}& \text{16384}\\ \text{2}& \text{8192}\\ \text{2}& \text{4096}\\ \text{2}& \text{2048}\\ \text{2}& \text{1024}\\ \text{2}& \text{512}\\ \text{2}& \text{256}\\ \text{2}& \text{128}\\ \text{2}& \text{64}\\ \text{2}& \text{32}\\ \text{2}& \text{16}\\ \text{2}& \text{8}\\ \text{2}& \text{4}\\ \text{2}& \text{2}\\ & \text{1}\end{array}$

$\sqrt[3]{3-32768}=-\sqrt[3]{332768}[\because \sqrt[3]{3-x}=-\sqrt[3]{3x}]$

Resolving $32768$ into prime factors
$32768=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

Now, by grouping the factors into the triplets of equal factors, we get,
$32768=(2\times 2\times 2)\times (2\times 2\times 2)\times (2\times 2\times 2)\times (2\times 2\times 2)\times (2\times 2\times 2)$

$\therefore \sqrt[3]{332768}=2\times 2\times 2\times 2\times 2$
$=32$
$\therefore \sqrt[3]{3-32768}=-32$

Hence, the cube root of $-32768$ is $-32$.