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 $\text {ind}$
 A: 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.
A: 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.
A: For all $x,y\in\mathbb{Z}$, if $x\equiv y\pmod{17}$, then $x^6\equiv y^6\pmod{17}$, so it's enough to check one number from each residue class. Take the residue classes $0,1\ldots,16,17$. We obtain:
$$a\in\{0,1,2,4,9,13,15,16\}$$

There is a quick way to find out how many possible values of $a$ there are. Since $17$ is prime, there exists a primitive root modulo $17$, say $r$. Now, suppose we have some $a,b\in\mathbb{N}$ with:
$$(r^a)^6\equiv (r^b)^6\pmod 17$$
or, using Fermat's little theorem:
$$16\mid 6(a-b)\iff8\mid a-b$$
It follows that there are $8$ possible residues.
A: The multiplicative group of the residues $\not = 0$ is cyclic. So there exists a generating element (called a primitive root) 
$\alpha.$ So the requested numbers are 
$$\{\alpha^2,\alpha^4,\alpha^6,\alpha^8,\alpha^{10},\alpha^{12},\alpha^{14},\alpha^{16}\}.$$
The following residues are primitive roots 
$$\{\alpha,\alpha^3,\alpha^5,\alpha^7,\alpha^{9},\alpha^{11},\alpha^{13},\alpha^{15}\}.$$ So if you select an arbitrary residue different from $0$ you have a chance of $0.5$ that it is a primitive root. A selected element is a primitive root if $\alpha^8 \not=1$. Actually it is sufficient to find $\alpha^2$. This is a number $\beta$ such that $b^4\not=1$ but $b^8=1.$ The residue $2$ is such a $\beta$, so 
$$2, 4, 8, 16 \;(=-1), -2, -4, -8,-16\;(=1)$$ are the requested numbers.
