Proving U(55) = U(55)^3 Algebraically

[e(ho0n3]

Homework Statement
Show that $U(55)^3 = \{x^3 \, : \, x \in U(55)\}$ is U(55).

The attempt at a solution
I wrote a Perl script that computes both U(55) and $U(55)^3$ and they both are equal. However, I want to solve this algebraically. My attempt is to define $f(x) = x^3$ (mod 55) and demonstrate that f is an automorphism. If f is an automorphism, surely U(55) = $U(55)^3$. I'm having trouble showing that f is injective.

Suppose f(a) = f(b) and $a \ne b$ where a, b are in U(55). f(a) = f(b) is equiv. to $a^3 = b^3$ (mod 55). I need is reduce this latter equation to a = b (mod 55) but I'm unable to do so.

 

[HallsofIvy]

Saying a³= b³ (mod 55) is the same as saying a³- b³= (a- b)(a²+ ab+ b²)= 0 (mod 55). Does U₅₅ have zero-divisors? What can you say about a²+ ab+ b²= 0 (mod 55)?

 

[e(ho0n3]

I would say U(55) doesn't have zero divisors for if it did, it would not be a group because of closure. If $a^2 + ab + b^2$ = 0 (mod 55), then $a^2 + ab + b^2$ is a multiple of 55 so it is not in U(55).