Solve for x 2 cube root of x+16=0

$2 \sqrt[3]{x} + 16 = 0$

Subtract $16$ from both sides of the equation.

$2 \sqrt[3]{x} = - 16$

To remove the radical on the left side of the equation, cube both sides of the equation.

${(2 \sqrt[3]{x})}^{3} = {(- 16)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

${(2 x^{\frac{1}{3}})}^{3} = {(- 16)}^{3}$

Simplify the left side.

Simplify ${(2 x^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $2 x^{\frac{1}{3}}$ .

$2^{3} {(x^{\frac{1}{3}})}^{3} = {(- 16)}^{3}$

Raise $2$ to the power of $3$ .

$8 {(x^{\frac{1}{3}})}^{3} = {(- 16)}^{3}$

Multiply the exponents in ${(x^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 x^{\frac{1}{3} \cdot 3} = {(- 16)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 x^{\frac{1}{3} \cdot 3} = {(- 16)}^{3}$

Rewrite the expression.

$8 x^{1} = {(- 16)}^{3}$

Simplify.

$8 x = {(- 16)}^{3}$

Simplify the right side.

Raise $- 16$ to the power of $3$ .

$8 x = - 4096$

Divide each term in $8 x = - 4096$ by $8$ and simplify.

Divide each term in $8 x = - 4096$ by $8$ .

$\frac{8 x}{8} = \frac{- 4096}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{- 4096}{8}$

Divide $x$ by $1$ .

$x = \frac{- 4096}{8}$

Simplify the right side.

Divide $- 4096$ by $8$ .

$x = - 512$