Pre-Algebra Examples

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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Step 2.1

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

$8 {({(x - 1)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.2.1.3

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

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Step 2.2.1.3.1

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

$8 {(x - 1)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.2.1.3.2

Cancel the common factor of $3$ .

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Step 2.2.1.3.2.1

Cancel the common factor.

$8 {(x - 1)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.2.1.3.2.2

Rewrite the expression.

$8 {(x - 1)}^{1} = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.2.1.4

Simplify.

$8 (x - 1) = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.2.1.5

Apply the distributive property.

$8 x + 8 \cdot - 1 = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.2.1.6

Multiply $8$ by $- 1$ .

$8 x - 8 = {\sqrt[3]{x^{2} + 2 x}}^{3}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify ${\sqrt[3]{x^{2} + 2 x}}^{3}$ .

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Step 2.3.1.1

Factor $x$ out of $x^{2} + 2 x$ .

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Step 2.3.1.1.1

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

$8 x - 8 = {\sqrt[3]{x \cdot x + 2 x}}^{3}$

Step 2.3.1.1.2

Factor $x$ out of $2 x$ .

$8 x - 8 = {\sqrt[3]{x \cdot x + x \cdot 2}}^{3}$

Step 2.3.1.1.3

Factor $x$ out of $x \cdot x + x \cdot 2$ .

$8 x - 8 = {\sqrt[3]{x (x + 2)}}^{3}$

Step 2.3.1.2

Rewrite ${\sqrt[3]{x (x + 2)}}^{3}$ as $x (x + 2)$ .

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Step 2.3.1.2.1

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

$8 x - 8 = {({(x (x + 2))}^{\frac{1}{3}})}^{3}$

Step 2.3.1.2.2

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

$8 x - 8 = {(x (x + 2))}^{\frac{1}{3} \cdot 3}$

Step 2.3.1.2.3

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

$8 x - 8 = {(x (x + 2))}^{\frac{3}{3}}$

Step 2.3.1.2.4

Cancel the common factor of $3$ .

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Step 2.3.1.2.4.1

Cancel the common factor.

$8 x - 8 = {(x (x + 2))}^{\frac{3}{3}}$

Step 2.3.1.2.4.2

Rewrite the expression.

$8 x - 8 = {(x (x + 2))}^{1}$

Step 2.3.1.2.5

Simplify.

$8 x - 8 = x (x + 2)$

Step 2.3.1.3

Apply the distributive property.

$8 x - 8 = x \cdot x + x \cdot 2$

Step 2.3.1.4

Simplify the expression.

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Step 2.3.1.4.1

Multiply $x$ by $x$ .

$8 x - 8 = x^{2} + x \cdot 2$

Step 2.3.1.4.2

Move $2$ to the left of $x$ .

$8 x - 8 = x^{2} + 2 x$

Step 3

Solve for $x$ .

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Step 3.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 2 x = 8 x - 8$

Step 3.2

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

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Step 3.2.1

Subtract $8 x$ from both sides of the equation.

$x^{2} + 2 x - 8 x = - 8$

Step 3.2.2

Subtract $8 x$ from $2 x$ .

$x^{2} - 6 x = - 8$

Step 3.3

Add $8$ to both sides of the equation.

$x^{2} - 6 x + 8 = 0$

Step 3.4

Factor $x^{2} - 6 x + 8$ using the AC method.

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Step 3.4.1

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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 3.4.2

Write the factored form using these integers.

$(x - 4) (x - 2) = 0$

Step 3.5

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

$x - 4 = 0$

$x - 2 = 0$

Step 3.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 3.6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.7

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 3.7.1

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

$x - 2 = 0$

Step 3.7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 3.8

The final solution is all the values that make $(x - 4) (x - 2) = 0$ true.

$x = 4, 2$