Question

Which of the following numbers are not perfect cubes?
$\left(1\right)216,\left(2\right)128,\left(3\right)1000,\left(4\right)100,\left(5\right)46656$

Answer 1

$\left(i\right)216$
$216=2\times 2\times 2\times 3\times 3\times 3$
$\sqrt[3]{3216}=\sqrt{2\times 2\times 2\times 3\times 3\times 3}$
$=2\times 3$
$=6$
$\left(ii\right)128$
$128=2\times 2\times 2\times 2\times 2\times 2\times 2$
$\sqrt[3]{3128}=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2}$
$=2\times 2\times \sqrt[3]{32}$
$=4\sqrt[3]{32}$
So, this is not a perfect cube.
$\left(iii\right)1000$
$1000=2\times 2\times 2\times 5\times 5\times 5$
$\sqrt[3]{31000}=\sqrt{2\times 2\times 2\times 5\times 5\times 5}$
$=2\times 5$
$=10$
$\left(iv\right)100$
$100=2\times 2\times 5\times 5$
$\sqrt[3]{3100}=\sqrt{2\times 2\times 5\times 5}$
So, this is not a perfect cube.
$\left(v\right)46656$
$46656=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3$
$\sqrt[3]{346656}=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3}$
$=2\times 2\times 3\times 3$
$=24$