Solve for x (7x-23)^(1/3)=(2x+17)^(1/3)

${(7 x - 23)}^{\frac{1}{3}} = {(2 x + 17)}^{\frac{1}{3}}$

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$7 x - 23 = 2 x + 17$

Solve for $x$ .

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

Subtract $2 x$ from both sides of the equation.

$7 x - 23 - 2 x = 17$

Subtract $2 x$ from $7 x$ .

$5 x - 23 = 17$

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

Add $23$ to both sides of the equation.

$5 x = 17 + 23$

Add $17$ and $23$ .

$5 x = 40$

Divide each term in $5 x = 40$ by $5$ and simplify.

Divide each term in $5 x = 40$ by $5$ .

$\frac{5 x}{5} = \frac{40}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{40}{5}$

Divide $x$ by $1$ .

$x = \frac{40}{5}$

Simplify the right side.

Divide $40$ by $5$ .

$x = 8$