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 $\text {ind}$

The group of nonzero residues modulo $17$ is not merely cyclic, it is cyclic of order $16$, which is relatively prime to $3$. This means that every residue has a unique cube root. As you very perceptively recognized, solving $x^6=a$ is the same as solving $y^2=a$; but you didn’t realize that given such a $y$, there is exactly one $x$ with $x^3=y$.
It follows, as @lhf pointed out in a comment, that the sixth powers in $\Bbb F_{17}^*$ are exactly the squares in this group.
