Simplify cube root of 16x^4- cube root of 128x

$\sqrt[3]{16 x^{4}} - \sqrt[3]{128 x}$

Step 1

Rewrite $16 x^{4}$ as ${(2 x)}^{3} \cdot (2 x)$ .

Factor $8$ out of $16$ .

$\sqrt[3]{8 (2) x^{4}} - \sqrt[3]{128 x}$

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

$\sqrt[3]{2^{3} \cdot 2 x^{4}} - \sqrt[3]{128 x}$

Factor out $x^{3}$ .

$\sqrt[3]{2^{3} \cdot 2 (x^{3} x)} - \sqrt[3]{128 x}$

Move $2$ .

$\sqrt[3]{2^{3} x^{3} \cdot 2 x} - \sqrt[3]{128 x}$

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

$\sqrt[3]{{(2 x)}^{3} \cdot 2 x} - \sqrt[3]{128 x}$

Add parentheses.

$\sqrt[3]{{(2 x)}^{3} \cdot (2 x)} - \sqrt[3]{128 x}$

Step 2

Pull terms out from under the radical.

$2 x \sqrt[3]{2 x} - \sqrt[3]{128 x}$

Step 3

Rewrite $128 x$ as $4^{3} \cdot (2 x)$ .

Factor $64$ out of $128$ .

$2 x \sqrt[3]{2 x} - \sqrt[3]{64 (2) x}$

Rewrite $64$ as $4^{3}$ .

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

Add parentheses.

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

Step 4

Pull terms out from under the radical.

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

Step 5

Multiply $4$ by $- 1$ .

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