Find all $x^6 \pmod {17}$ 
Find all $a$ such that $x^6 \equiv a\pmod {17}$ (not including $0$)

First I thought that we could look at $y^2 \equiv a \pmod {17}$, where $y=x^3$.
Then, by Euler's criterion, it must be that:
$$a^{\frac{17 -1}{2}} \equiv a^8 \equiv 1 \pmod {17}$$
I could develop it to
$$a^8 -1 \equiv 0 \pmod {17} \implies\\ (a-1)(a+1)(a^2+1)(a^4+1)\equiv 0\pmod {17}$$
I'm not sure it's the right way. Could you guide me please?
By the way, we are not familiar with $\text {ind}$

I got $$a\in\{0,1,2,4,8,9,13,15,16\}$$ and it occurs for
$$x\in\{0,1,5,8,6,7,2,3,4\}$$ respectively.
