Question

Find the cube root of $27\times 64$.

Answer 1

Resolving $27$ and $64$ into prime factors, we get
$\sqrt[3]{327}=(3\times 3\times 3{)}^{\frac{1}{3}}=(3^3{)}^{\frac{1}{3}}$

$\sqrt[3]{327}=3$
$\sqrt[3]{364}=(2\times 2\times 2\times 2\times 2\times 2{)}^{\frac{1}{3}}$

$=(2^6{)}^{\frac{1}{3}}=2^2=4$

$\sqrt[3]{363}=4$

Now, $\sqrt[3]{327\times 64}=\sqrt[3]{327}\times \sqrt[3]{364}$

$=3\times 4$

$\therefore \sqrt[3]{327\times 64}$ is the cube root of $12$