Solve for x 1=3+ cube root of 4x-8

$1 = 3 + \sqrt[3]{4 x - 8}$

Rewrite the equation as $3 + \sqrt[3]{4 x - 8} = 1$ .

$3 + \sqrt[3]{4 x - 8} = 1$

Move all terms not containing $\sqrt[3]{4 x - 8}$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$\sqrt[3]{4 x - 8} = 1 - 3$

Subtract $3$ from $1$ .

$\sqrt[3]{4 x - 8} = - 2$

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

${\sqrt[3]{4 x - 8}}^{3} = {(- 2)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 x - 8}$ as ${(4 x - 8)}^{\frac{1}{3}}$ .

${({(4 x - 8)}^{\frac{1}{3}})}^{3} = {(- 2)}^{3}$

Simplify the left side.

Simplify ${({(4 x - 8)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(4 x - 8)}^{\frac{1}{3}})}^{3}$ .

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

${(4 x - 8)}^{\frac{1}{3} \cdot 3} = {(- 2)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(4 x - 8)}^{\frac{1}{3} \cdot 3} = {(- 2)}^{3}$

Rewrite the expression.

${(4 x - 8)}^{1} = {(- 2)}^{3}$

Simplify.

$4 x - 8 = {(- 2)}^{3}$

Simplify the right side.

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

$4 x - 8 = - 8$

Solve for $x$ .

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

Add $8$ to both sides of the equation.

$4 x = - 8 + 8$

Add $- 8$ and $8$ .

$4 x = 0$

Divide each term in $4 x = 0$ by $4$ and simplify.

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$x = 0$

Solve for x 1=3+ cube root of 4x-8

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