Cubic residues in $Z_p^*$ I've the following question:
If prime $p \equiv 7 \bmod 9$ show that $x^{(p+2)/9}$ is a cubic root of $x$ in $\mathbb Z_{p}^*$, when it is known that $x$ is a cubic residue.
Basically, I've to show ${(x^{(p+2)/9})}^3 \equiv x \bmod p$. Can't seem to find a way to begin with. Clearly, $(p+2) \equiv 0 \bmod 9$. What does this really say ? 
 A: Hint $ $ It is just lil Fermat for a cube $x$, but scaled $\color{#c00}{ \times\, x}$. 
$$\begin{array}{rl}
0 \not\equiv x \equiv a^{\large 3}\,\ \overset{\rm Fermat}\Longrightarrow\!\!\! 
   &\! 1\equiv \overbrace{x^{\Large\color{#0a0}{ \frac{p-1}3}}}^{\Large a^{\Large p-1}} \\
\overset{\Large\color{#c00}{\times\, x}}\Longrightarrow 
   &\! x\equiv x^{\Large\frac{p+2}3}\!\!\equiv\! {\huge[} x^{\Large\color{#a0f}{\frac{p+2}9}}{\huge ]}^{\large 3}\ \ {\rm by}\,\ \ \begin{align} &9\mid p\!-\!7\\ \Rightarrow\ &\color{#a0f}{9\mid p\!+\!2}\\ \Rightarrow\ &\color{#0a0}{3\mid p\!-\!1}\end{align}
\end{array}$$

Remark $\ $ It easily generalizes from $\,3\to k$ 'th powers: $ $ if $\,p\equiv k^2\!+\!1\!-\!k\pmod{k^2}\ $ then 
$ 0\not\equiv x \equiv a^{\large k}\,\overset{\rm Fermat}\Longrightarrow1\equiv \underbrace{x^{\!\Large \color{#0a0}{\frac{p-1}k}}}_{\Large a^{\Large p-1}}$ $ \overset{\color{#c00}{\Large \times\, x}}\Longrightarrow \,x\equiv x^{\!\Large\frac{p+k-1}k}\!\!\equiv\! {\Huge [} x^{\!\color{#a0f}{\Large\frac{p+k-1}{k^2}}}{\!\Huge ]}^{\large k}$ ${\rm by}\ \   \begin{align} &k^2\mid p\!+\!k\!-\!1\!-\!k^2\\ \Rightarrow\ &\color{#a0f}{k^2\mid p\!+\!k\!-\!1}\\ \Rightarrow\ &\,\ \color{#0a0}{k\mid p\!-\!1} \end{align}  $
A: Assuming $p$ is prime. If $x\equiv a^3$, then
$$
\left(x^{\frac{p+2}{9}}\right)^3
\equiv \left(a^{\frac{p+2}{3}}\right)^3
\equiv a^{p+2}
\equiv a^{p-1}a^3
\equiv x
$$
A: Since we are given that $x$ is a cubic residue$\bmod p$ prime, we know that, given a primitive root $g$ of $p,$ we can find a $k$ such that $x \equiv g^{3k} \bmod p$ (since the cube roots of $x$ must be equivalent to some power of $g$).
Then define $a := x^{(p+2)/9} \equiv (g^{3k})^{(p+2)/9} \equiv g^{k(p+2)/3} \equiv g^{k+k(p-1)/3}\bmod p$.
Now we have $a^3 \equiv  g^{3k+k(p-1)} \equiv g^{3k} \equiv x \bmod p$ as required.
A: Let $x\ne 0$ in $\Bbb F_p$. One has 
$$x^{\frac{p+2}{9}}=x^{\frac13}\iff x^{\frac{p+2}{3}}=x\iff x^{\frac {p-1}{3}}=1 $$
 Since $x$ is a cube in $\Bbb F_p$ we have simply the Fermat's Little theorem.
(The restriction for the prime $p$ is not necessary).
