Simplify cube root of 40x^8

$\sqrt[3]{40 x^{8}}$

Step 1

Rewrite $40 x^{8}$ as ${(2 x^{2})}^{3} \cdot (5 x^{2})$ .

Factor $8$ out of $40$ .

$\sqrt[3]{8 (5) x^{8}}$

Rewrite $8$ as $2^{3}$ .

$\sqrt[3]{2^{3} \cdot 5 x^{8}}$

Factor out $x^{6}$ .

$\sqrt[3]{2^{3} \cdot 5 (x^{6} x^{2})}$

Rewrite $x^{6}$ as ${(x^{2})}^{3}$ .

$\sqrt[3]{2^{3} \cdot 5 ({(x^{2})}^{3} x^{2})}$

Move $5$ .

$\sqrt[3]{2^{3} {(x^{2})}^{3} \cdot 5 x^{2}}$

Rewrite $2^{3} {(x^{2})}^{3}$ as ${(2 x^{2})}^{3}$ .

$\sqrt[3]{{(2 x^{2})}^{3} \cdot 5 x^{2}}$

Add parentheses.

$\sqrt[3]{{(2 x^{2})}^{3} \cdot (5 x^{2})}$

Step 2

Pull terms out from under the radical.

$2 x^{2} \sqrt[3]{5 x^{2}}$