Solve by Factoring cube root of 5x-4 = cube root of 7x+8

$\sqrt[3]{5 x - 4} = \sqrt[3]{7 x + 8}$

Step 1

Subtract $\sqrt[3]{7 x + 8}$ from both sides of the equation.

$\sqrt[3]{5 x - 4} - \sqrt[3]{7 x + 8} = 0$

Step 2

Add $\sqrt[3]{7 x + 8}$ to both sides of the equation.

$\sqrt[3]{5 x - 4} = \sqrt[3]{7 x + 8}$

Step 3

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

${\sqrt[3]{5 x - 4}}^{3} = {\sqrt[3]{7 x + 8}}^{3}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{5 x - 4}$ as ${(5 x - 4)}^{\frac{1}{3}}$ .

${({(5 x - 4)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{7 x + 8}}^{3}$

Simplify the left side.

Simplify ${({(5 x - 4)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(5 x - 4)}^{\frac{1}{3}})}^{3}$ .

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

${(5 x - 4)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{7 x + 8}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(5 x - 4)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{7 x + 8}}^{3}$

Rewrite the expression.

${(5 x - 4)}^{1} = {\sqrt[3]{7 x + 8}}^{3}$

Simplify.

$5 x - 4 = {\sqrt[3]{7 x + 8}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{7 x + 8}}^{3}$ as $7 x + 8$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7 x + 8}$ as ${(7 x + 8)}^{\frac{1}{3}}$ .

$5 x - 4 = {({(7 x + 8)}^{\frac{1}{3}})}^{3}$

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

$5 x - 4 = {(7 x + 8)}^{\frac{1}{3} \cdot 3}$

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

$5 x - 4 = {(7 x + 8)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$5 x - 4 = {(7 x + 8)}^{\frac{3}{3}}$

Rewrite the expression.

$5 x - 4 = {(7 x + 8)}^{1}$

Simplify.

$5 x - 4 = 7 x + 8$

Step 5

Solve for $x$ .

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

Subtract $7 x$ from both sides of the equation.

$5 x - 4 - 7 x = 8$

Subtract $7 x$ from $5 x$ .

$- 2 x - 4 = 8$

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$- 2 x = 8 + 4$

Add $8$ and $4$ .

$- 2 x = 12$

Divide each term in $- 2 x = 12$ by $- 2$ and simplify.

Divide each term in $- 2 x = 12$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{12}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{12}{- 2}$

Divide $x$ by $1$ .

$x = \frac{12}{- 2}$

Simplify the right side.

Divide $12$ by $- 2$ .

$x = - 6$

Solve by Factoring cube root of 5x-4 = cube root of 7x+8

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