Solve for x cube root of x+1 = cube root of x^2-5

$\sqrt[3]{x + 1} = \sqrt[3]{x^{2} - 5}$

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

${\sqrt[3]{x + 1}}^{3} = {\sqrt[3]{x^{2} - 5}}^{3}$

Simplify each side of the equation.

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

${({(x + 1)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{x^{2} - 5}}^{3}$

Simplify the left side.

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

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

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

${(x + 1)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{x^{2} - 5}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x + 1)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{x^{2} - 5}}^{3}$

Rewrite the expression.

${(x + 1)}^{1} = {\sqrt[3]{x^{2} - 5}}^{3}$

Simplify.

$x + 1 = {\sqrt[3]{x^{2} - 5}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{x^{2} - 5}}^{3}$ as $x^{2} - 5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{2} - 5}$ as ${(x^{2} - 5)}^{\frac{1}{3}}$ .

$x + 1 = {({(x^{2} - 5)}^{\frac{1}{3}})}^{3}$

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

$x + 1 = {(x^{2} - 5)}^{\frac{1}{3} \cdot 3}$

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

$x + 1 = {(x^{2} - 5)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x + 1 = {(x^{2} - 5)}^{\frac{3}{3}}$

Rewrite the expression.

$x + 1 = {(x^{2} - 5)}^{1}$

Simplify.

$x + 1 = x^{2} - 5$

Solve for $x$ .

Subtract $x^{2}$ from both sides of the equation.

$x + 1 - x^{2} = - 5$

Add $5$ to both sides of the equation.

$x + 1 - x^{2} + 5 = 0$

Add $1$ and $5$ .

$x - x^{2} + 6 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 6$ .

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 6 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 6 = 0$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 6 = 0$

Rewrite $6$ as $- 1 (- 6)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 6)$ .

$- (x^{2} - x - 6) = 0$

Factor.

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$- ((x - 3) (x + 2)) = 0$

Remove unnecessary parentheses.

$- (x - 3) (x + 2) = 0$

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

$x - 3 = 0$

$x + 2 = 0$

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 $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $- (x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Solve for x cube root of x+1 = cube root of x^2-5

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