Simplify 7( cube root of 2x)-3 cube root of 16x-3 cube root of 8x

$7 (\sqrt[3]{2 x}) - 3 \sqrt[3]{16 x} - 3 \sqrt[3]{8 x}$

Step 1

Simplify each term.

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

Factor $8$ out of $16$ .

$7 \sqrt[3]{2 x} - 3 \sqrt[3]{8 (2) x} - 3 \sqrt[3]{8 x}$

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

$7 \sqrt[3]{2 x} - 3 \sqrt[3]{2^{3} \cdot 2 x} - 3 \sqrt[3]{8 x}$

Add parentheses.

$7 \sqrt[3]{2 x} - 3 \sqrt[3]{2^{3} \cdot (2 x)} - 3 \sqrt[3]{8 x}$

Pull terms out from under the radical.

$7 \sqrt[3]{2 x} - 3 (2 \sqrt[3]{2 x}) - 3 \sqrt[3]{8 x}$

Multiply $2$ by $- 3$ .

$7 \sqrt[3]{2 x} - 6 \sqrt[3]{2 x} - 3 \sqrt[3]{8 x}$

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

$7 \sqrt[3]{2 x} - 6 \sqrt[3]{2 x} - 3 \sqrt[3]{2^{3} x}$

Pull terms out from under the radical.

$7 \sqrt[3]{2 x} - 6 \sqrt[3]{2 x} - 3 (2 \sqrt[3]{x})$

Multiply $2$ by $- 3$ .

$7 \sqrt[3]{2 x} - 6 \sqrt[3]{2 x} - 6 \sqrt[3]{x}$

Step 2

Subtract $6 \sqrt[3]{2 x}$ from $7 \sqrt[3]{2 x}$ .

$\sqrt[3]{2 x} - 6 \sqrt[3]{x}$