Given $x\equiv b^{(2p-1)/3} \pmod{p}$, show that $x^{3} \equiv b \pmod{p}$. So I have a prime $p$, where $p \equiv 2 \pmod{3}$, and $b \not\equiv 0 \pmod{p}$. Given $x\equiv b^{(2p-1)/3} \pmod{p}$, I'm supposed to show that $x^{3} \equiv b \pmod{p}$.
I tried to use Fermat's little theorem for this, but to no avail. I also tried to say something along the lines of

Let $g$ be a primitive root, so that $g^{i} \equiv b$ for some $i$:
  $$1 \equiv b^{(2p-1)/3} \equiv g^{i(2p-1)/3} \pmod{p}$$
  Since the primitive root has order $p-1$, then $p-1 \mid i(2p-1)/3$ and $(p-1)k = i(2p-1)/3$ for some $k$.

From there I was hoping things would simplify nicely, but it doesn’t seem to. I figure I need to show that $i\equiv 0 \pmod{3}$ but I am not sure about doing that.
Apparently when $p \equiv 1 \pmod{3}$, showing that $i \equiv 0 \pmod{3}$ implies that $b$ is a cube mod $p$.
Thanks for any help.
 A: Start by cubing both sides to get:
$$x^3\equiv b^{2p-1} \pmod p$$
Now we can use FLT to get:
$$b^{2p-1}=b^{p-1}\cdot b^{p}\equiv1\cdot b \pmod p$$
EDIT: Note that you will also have to address the special case where $b\equiv 0\pmod p$, but that one is pretty trivial.
A: Intuitively it is simply $\,x\equiv b^k\!\equiv b^{\color{#0a0}{1/3}}$ $\Rightarrow$ $\,x^3\equiv b,\,$ by  $\ k\equiv\color{#0a0}{1/3}\pmod{\!p\!-\!1},\ $ i.e.
$\quad {\rm mod}\,\ p\!-\!1\!:\,\ \color{#c00}{p\equiv 1}\,\Rightarrow\, k = \dfrac{(2\,\color{#c00}p\!-\!1)}3\equiv \dfrac{2\cdot \color{#c00}1-1}3\equiv\color{#0a0}{\dfrac{1}3}\,\ $ so $\,\ b^{\large k}\equiv\, b^{\large\color{#0a0}{ 1/3}}$

Note exponents on $\,b\,$ can be taken $\,{\rm mod}\,\ \color{#c00}{p\!-\!1}\ $ by $\,b^{\large\color{#c00}{p-1}}\equiv 1\pmod p\ $ by  
Lemma $\ \ \color{#c00}{b^{\large n} = 1}\ $ and  $\ i\equiv j\pmod{n}\ \Rightarrow\ b^{\large i} = b^{\large j}$
Proof $\ \ i = j\! +\! kn\,\Rightarrow\, b^{\large i} = b^{\large j+kn} = b^{\large j}(\color{#c00}{b^{\large n}})^{\large k} = b^{\large j}\color{#c00}1^{\large k} = b^{\large j}$
