Solve by Factoring cube root of 7x-8-3=0

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

Step 1

Add $3$ to both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${({(7 x - 8)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(7 x - 8)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

${(7 x - 8)}^{1} = 3^{3}$

Simplify.

$7 x - 8 = 3^{3}$

Simplify the right side.

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

$7 x - 8 = 27$

Step 4

Solve for $x$ .

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

Add $8$ to both sides of the equation.

$7 x = 27 + 8$

Add $27$ and $8$ .

$7 x = 35$

Divide each term in $7 x = 35$ by $7$ and simplify.

Divide each term in $7 x = 35$ by $7$ .

$\frac{7 x}{7} = \frac{35}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{35}{7}$

Divide $x$ by $1$ .

$x = \frac{35}{7}$

Simplify the right side.

Divide $35$ by $7$ .

$x = 5$