## Intermediate Algebra for College Students (7th Edition)

$x = 11$
Subtract 5 on both sides to obtain: $(3x-6)^{\frac{1}{3}}=3$ Cube both sides to obtain: $\left((3x-6)^{\frac{1}{3}}\right)^3=3^3 \\(3x-6)^{\frac{1}{3}(3)}=27 \\(3x-6)^1=27 \\3x-6=27$ Aadd 6 on both sides to obtain: $3x=33$ Divide 3 on both sides to obtain: $x = 11$