Question

Find the cube roots of $1728$ by prime factorization:

Answer 1

On prime factorising, we get,

$1728=(2\times 2\times 2)\times (2\times 2\times 2)\times (3\times 3\times 3)$

$=2^3\times 2^3\times 3^3$.

Then, cube root of $1728$ is:

$\sqrt[3]{31728}=\sqrt[3]{32^3\times 2^3\times 3^3}=2\times 2\times 3=12$.

Therefore, $12$ is the solution.