Question

Simplify: $\frac{\sqrt[3]{3512}-\sqrt[3]{3216}}{\sqrt[3]{3125}-\sqrt[3]{364}}$.

Answer 1

On prime factorisation of the numbers individually, we get,

$512=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}$

$\Rightarrow \sqrt[3]{3512}=2\times 2\times 2=8$.

$216=\underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}$

$\Rightarrow \sqrt[3]{3216}=2\times 3=6$.

$125=\underset{–}{5\times 5\times 5}$

$\Rightarrow \sqrt[3]{3125}=5$.

$64=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}$

$\Rightarrow \sqrt[3]{364}=2\times 2=4$.

Therefore,

$\frac{\sqrt[3]{3512}-\sqrt[3]{3216}}{\sqrt[3]{3125}-\sqrt[3]{364}}=\frac{8-6}{5-4}=\frac{2}{1}=2$.

Thus, $2$ is the solution.