Solve for x cube root of x^2-x- cube root of x+99=0

$\sqrt[3]{x^{2} - x} - \sqrt[3]{x + 99} = 0$

Step 1

Add $\sqrt[3]{x + 99}$ to both sides of the equation.

$\sqrt[3]{x^{2} - x} = \sqrt[3]{x + 99}$

Step 2

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

${\sqrt[3]{x^{2} - x}}^{3} = {\sqrt[3]{x + 99}}^{3}$

Step 3

Simplify each side of the equation.

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

${({(x^{2} - x)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{x + 99}}^{3}$

Simplify the left side.

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

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

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

${(x^{2} - x)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{x + 99}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} - x)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{x + 99}}^{3}$

Rewrite the expression.

${(x^{2} - x)}^{1} = {\sqrt[3]{x + 99}}^{3}$

Simplify.

$x^{2} - x = {\sqrt[3]{x + 99}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{x + 99}}^{3}$ as $x + 99$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x + 99}$ as ${(x + 99)}^{\frac{1}{3}}$ .

$x^{2} - x = {({(x + 99)}^{\frac{1}{3}})}^{3}$

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

$x^{2} - x = {(x + 99)}^{\frac{1}{3} \cdot 3}$

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

$x^{2} - x = {(x + 99)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x^{2} - x = {(x + 99)}^{\frac{3}{3}}$

Rewrite the expression.

$x^{2} - x = {(x + 99)}^{1}$

Simplify.

$x^{2} - x = x + 99$

Step 4

Solve for $x$ .

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

Subtract $x$ from both sides of the equation.

$x^{2} - x - x = 99$

Subtract $x$ from $- x$ .

$x^{2} - 2 x = 99$

Subtract $99$ from both sides of the equation.

$x^{2} - 2 x - 99 = 0$

Factor $x^{2} - 2 x - 99$ 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 $- 99$ and whose sum is $- 2$ .

$- 11, 9$

Write the factored form using these integers.

$(x - 11) (x + 9) = 0$

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

$x - 11 = 0$

$x + 9 = 0$

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

The final solution is all the values that make $(x - 11) (x + 9) = 0$ true.

$x = 11, - 9$

Solve for x cube root of x^2-x- cube root of x+99=0

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