Simplify cube root of 2x^5* cube root of 64x^9

$\sqrt[3]{2 x^{5}} \cdot \sqrt[3]{64 x^{9}}$

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

Factor out $x^{3}$ .

$\sqrt[3]{2 (x^{3} x^{2})} \cdot \sqrt[3]{64 x^{9}}$

Reorder $2$ and $x^{3}$ .

$\sqrt[3]{x^{3} \cdot 2 x^{2}} \cdot \sqrt[3]{64 x^{9}}$

Add parentheses.

$\sqrt[3]{x^{3} \cdot (2 x^{2})} \cdot \sqrt[3]{64 x^{9}}$

Pull terms out from under the radical.

$x \sqrt[3]{2 x^{2}} \cdot \sqrt[3]{64 x^{9}}$

Rewrite $64 x^{9}$ as ${(4 x^{3})}^{3}$ .

$x \sqrt[3]{2 x^{2}} \cdot \sqrt[3]{{(4 x^{3})}^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x \sqrt[3]{2 x^{2}} \cdot (4 x^{3})$

Rewrite using the commutative property of multiplication.

$4 x \sqrt[3]{2 x^{2}} x^{3}$

Multiply $x$ by $x^{3}$ by adding the exponents.

Move $x^{3}$ .

$4 (x^{3} x) \sqrt[3]{2 x^{2}}$

Multiply $x^{3}$ by $x$ .

Raise $x$ to the power of $1$ .

$4 (x^{3} x^{1}) \sqrt[3]{2 x^{2}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 x^{3 + 1} \sqrt[3]{2 x^{2}}$

Add $3$ and $1$ .

$4 x^{4} \sqrt[3]{2 x^{2}}$