(Cube root of x - 2)^3-3

I don't know how to write it, but.. help :(

Does the ^3 cancel out the cube root? So it would be x-5?

Depends.

Is all of (x-2) under cube root or only x?
In the first case yes, in the second case no.

Case 1:

(∛(x-2))³ = ((x-2)¹/³)³ = (x-2)¹/³*³ = x - 2
(∛(x-2))³ - 3 = x - 2 - 3 = x - 5

Case 2:

((∛x) - 2)³ = (∛x)³(-2)⁰ + 3 (∛x)²(-2)¹ + 3 (∛x)¹(-2)² + (∛x)⁰(-2)³
((∛x) - 2)³ = x - 6 x²/³ + 12 x¹/³ - 8

((∛x) - 2)³ - 3 = x - 6 x²/³ + 12 x¹/³ - 8 - 3 = x - 6 x²/³ + 12 x¹/³ - 11

Yes the answer is x -5