Proving that the cubic operation is injective

Question

Consider $\mathbb Z_{17}$. I want to prove that if $x^3\equiv y^3 \pmod{17}$ then $x\equiv y \pmod{17}$.

Here's a start:

$$x^3 \equiv y^3 \pmod{17} \\ \left( \frac{x}{y} \right)^3 \equiv 1 \pmod{17}$$

How can I deduce that $x\equiv y \pmod{17}$?

Thanks

EDIT
I think I know. Let's denote $r := ord_{17} (\frac{x}{y})$ then $r \mid 3$ and $r \mid 16$ (by Fermmat's theorem) but $\gcd(3,16)=1$ so $r=1$. so $\frac{x}{y} = 1 \implies x\equiv y \pmod{17}$.

Answer 1

By Fermat's little theorem, if $x\not\equiv0\pmod{17}$, then $x^{16}\equiv1\pmod{17}$.

If $x^3\equiv1\pmod{17}$, then $x^{16}\equiv x\cdot (x^3)^5\equiv x\equiv1\pmod{17}$. So if $x^3\equiv1\pmod{17}$, then $x\equiv1\pmod{17}$

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Hope this helps.