Solve for x cube root of x^2+2x-12 = cube root of -2x+9

$\sqrt[3]{x^{2} + 2 x - 12} = \sqrt[3]{- 2 x + 9}$

Step 1

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

${\sqrt[3]{x^{2} + 2 x - 12}}^{3} = {\sqrt[3]{- 2 x + 9}}^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{2} + 2 x - 12}$ as ${(x^{2} + 2 x - 12)}^{\frac{1}{3}}$ .

${({(x^{2} + 2 x - 12)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{- 2 x + 9}}^{3}$

Simplify the left side.

Simplify ${({(x^{2} + 2 x - 12)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x^{2} + 2 x - 12)}^{\frac{1}{3}})}^{3}$ .

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

${(x^{2} + 2 x - 12)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{- 2 x + 9}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} + 2 x - 12)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{- 2 x + 9}}^{3}$

Rewrite the expression.

${(x^{2} + 2 x - 12)}^{1} = {\sqrt[3]{- 2 x + 9}}^{3}$

Simplify.

$x^{2} + 2 x - 12 = {\sqrt[3]{- 2 x + 9}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{- 2 x + 9}}^{3}$ as $- 2 x + 9$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{- 2 x + 9}$ as ${(- 2 x + 9)}^{\frac{1}{3}}$ .

$x^{2} + 2 x - 12 = {({(- 2 x + 9)}^{\frac{1}{3}})}^{3}$

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

$x^{2} + 2 x - 12 = {(- 2 x + 9)}^{\frac{1}{3} \cdot 3}$

Combine $\frac{1}{3}$ and $3$ .

$x^{2} + 2 x - 12 = {(- 2 x + 9)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x^{2} + 2 x - 12 = {(- 2 x + 9)}^{\frac{3}{3}}$

Rewrite the expression.

$x^{2} + 2 x - 12 = {(- 2 x + 9)}^{1}$

Simplify.

$x^{2} + 2 x - 12 = - 2 x + 9$

Step 3

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Add $2 x$ to both sides of the equation.

$x^{2} + 2 x - 12 + 2 x = 9$

Add $2 x$ and $2 x$ .

$x^{2} + 4 x - 12 = 9$

Subtract $9$ from both sides of the equation.

$x^{2} + 4 x - 12 - 9 = 0$

Subtract $9$ from $- 12$ .

$x^{2} + 4 x - 21 = 0$

Factor $x^{2} + 4 x - 21$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

$(x - 3) (x + 7) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 7 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 3) (x + 7) = 0$ true.

$x = 3, - 7$

Solve for x cube root of x^2+2x-12 = cube root of -2x+9

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