Question

Is $128$ a perfect cube?Find its cube root.

Answer 1

$1252418$

$\therefore$ $128=(2\times 2\times 2)\times (2\times 2\times 2)\times 2$

$\sqrt[3]{3128}={(2^3\times 2^3\times 2^{})}^{1/3}=2^{3/3}\times 2^{2/3}\times 2^{1/3}=2\times 2\times \sqrt[3]{32}=4\sqrt[3]{32}$

$\therefore$ $128$ is not a perfect cube

Its cube root is $4\sqrt[3]{32}$