Solve for x cube root of x^2+2x=-1

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

Step 1

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

${\sqrt[3]{x^{2} + 2 x}}^{3} = {(- 1)}^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{2} + 2 x}$ as ${(x^{2} + 2 x)}^{\frac{1}{3}}$ .

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

Simplify the left side.

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

Multiply the exponents in ${({(x^{2} + 2 x)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

${(x^{2} + 2 x)}^{1} = {(- 1)}^{3}$

Simplify.

$x^{2} + 2 x = {(- 1)}^{3}$

Simplify the right side.

Raise $- 1$ to the power of $3$ .

$x^{2} + 2 x = - 1$

Step 3

Solve for $x$ .

Add $1$ to both sides of the equation.

$x^{2} + 2 x + 1 = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x^{2} + 2 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$