Find values of $p$ such that $p^2-p+1$ is cubic? Find all prime numbers such that $p^2-p+1$ is a cubic number. I've tried substituting values of $x$ for $p^2-p+1 = x^3$. 
This gave me a few values, which were small. As values of $x$ ranged higher and higher, it was tedious to find values of $p$ that satisfied this condition. To my imminent horror, I realized that most of these values of $p$ where in fact NOT prime numbers. 
Can someone guide me through the solution, as I prefer to use a slick solution, rather than using brute force? After the solution, can someone point me towards resources for solving problems like this?

Here is a solution using simpler tools.
Our starting point is
$$p^2 - p + 1 = x^3$$
with $p$ prime, and $x$ is an integer.
As the smallest prime is 2, note that 
$$x^3 = p^2 - p + 1 \ge 2^2 - 2 + 1 = 3 \quad \rightarrow \quad x \ge 3^{1/3} > 1$$
$$p^2 > p^2 - p + 1 = x^3 \quad \rightarrow \quad p > x^{3/2} > x $$
Since $p$ being prime is important, this hints we should try to rearrange the equation to factor as much as possible.
$$
\begin{eqnarray}
p^2 - p + 1 &=& x^3 \\
p^2 - p &=& x^3 - 1 \\
p(p-1) &=& (x-1)(x^2+x+1)
\end{eqnarray}
$$
Because $p$ is prime and divides the left hand side, it must divide the right hand side as well. Since $p>x$, this means $p$ cannot divide $(x-1)$ and therefore $(x^2+x+1)$ must be a multiple of $p$. Let's call this multiple $m$. 
So we now have two equations
$$
\begin{eqnarray}
mp &=& (x^2+x+1) \\
(p-1) &=& m(x-1)
\end{eqnarray}
$$
Because $p>x$ the second equation makes it clear that $m>1$, and since it is an integer $m \ge 2$. Note also that $(x^2 + x +1)$ cannot be even, thus raising the limit further: $m \ge 3$.
Eliminating $p$ to get a quadratic on $x$
$$
\begin{eqnarray}
p &=& mx - m + 1 \\
m(mx - m + 1) &=& (x^2+x+1) \\
0 &=& x^2 + (1-m^2)x + (m^2-m+1)
\end{eqnarray}
$$
For the quadratic equation to have an integer solution, the discriminant must be a square value
$$
\begin{eqnarray}
(1-m^2)^2 - 4(m^2-m+1) &=& (1 - 2m^2 + m^4) - 4(m^2-m+1) \\
                       &=& m^4 - 6m^2 +4m-3
\end{eqnarray}
$$
Requiring this to be a square integer is quite restrictive, which is easier to see if we denote the square as $(m^2 - k)^2 = m^4 -2km^2 + k^2$. Using this along with the fact that $m \ge 3$ (thus $m^2 \ge 3m$), we can see
$$ m^4 - 6m^2 +4m-3 < m^4 - 4m^2 - 3 < (m^2 - 2)^2 $$
and
$$ m^4 - 6m^2 +4m-3 \ge m^4 - 6m^2 -3m + 18 > m^4 -8m^2 + 18 > (m^2 - 4)^2 $$ 
therefore
$$m^4 - 6m^2 +4m-3 = (m^2 - 3)^2 = m^4 - 6m^2 + 9$$
$$4m = 12 \quad \rightarrow \quad m=3$$
Now putting this back in our quadratic equation for $x$
$$ 0 = x^2 + (1-m^2)x + (m^2-m+1) = x^2 - 8x + 7 = (x-1)(x-7) $$
As $x>1$, this means the only solution is
$$x=7 \quad \rightarrow \quad p=m(x-1)+1=19. $$
