f(x)=4-2(x-6)^1/3 evaluate when f(x)=5

Question

so I did this:

5=4-2(x-6)^1/3
1=-2(x-6)^1/3
(-1/2)=(x-6)^1/3

How do I get rid of the exponent 1/3? I am stumped.

Answer 1

cube both sides
(an exponent of 1/3 means cube root, so what is the inverse of taking the cube root ?)

(-1/2)^3=[(x-6)^1/3]^3
-1/8 = x-6
x = 6-1/8
x = 47/8