Simplify cube root of 162x^6y^7

$\sqrt[3]{162 x^{6} y^{7}}$

Step 1

Rewrite $162 x^{6} y^{7}$ as ${(3 x^{2} y^{2})}^{3} \cdot (6 y)$ .

Factor $27$ out of $162$ .

$\sqrt[3]{27 (6) x^{6} y^{7}}$

Rewrite $27$ as $3^{3}$ .

$\sqrt[3]{3^{3} \cdot 6 x^{6} y^{7}}$

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

$\sqrt[3]{3^{3} \cdot 6 {(x^{2})}^{3} y^{7}}$

Factor out $y^{6}$ .

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

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

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

Move $6$ .

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

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

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

Add parentheses.

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

Step 2

Pull terms out from under the radical.

$3 x^{2} y^{2} \sqrt[3]{6 y}$