At a first glance, I would do a PBC and something like $p^2+p+1=n^3$, which is $p^3-1=(p-1)n^3$.

The same idea used in the BMO 2005

why $$ k^4 >k^2 -6k^2 - 4k - 3 >(k^2-4)^2 $$

Kimchiks926 wrote: Let $p$ be a prime. Prove that $p^2+p+1$ is never a perfect cube. Sorry there was some calculation error in my solution...

abhijeetkr wrote: Kimchiks926 wrote: Let $p$ be a prime. Prove that $p^2+p+1$ is never a perfect cube. I think a smaller soln. Could be to use the fact that any prime >2 is odd....We can replace it as 2k+1, we could put it in the eqn. And then factorise it.....we would get (2k+1)(2k+3)....which would never be a perfect cube as 2k+1 is prime. Another case could be p = 2, at that instant the eqn. will give 11 which is not a perfect cube.... But $(2k+1)(2k+3)$ is not equal to $(2k+1)^2+(2k+1)+1$

Since a lot of people do not understand or don't like my quite technical and a bit bashy solution, which I found during the contest, then I will present another solution involving quadratic residues. Assume that $p^2+p+1 = a^3 \implies p(p+1) = (a-1)(a^2 +a+1)$. If $p \mid a-1$, we can proceed similarly as I did in previous solution. Now we move on to the case $p \mid a^2 +a+1 $. If $p=3$, then $a^3 =13 $, which is clearly impossible. If $p \equiv 1 \pmod 3$, we can assume that $p=3k+1$ and consequently we have: $$ p^2+p+ 1 =9k^2 + 9k +3 \equiv 3 \pmod 9 $$but no perfect cube is $3 \pmod 9$ IF $p \equiv 2 \pmod 3$, we use the fact that: $$ a^2 +a + 1 \equiv 0 \pmod p \implies (2a+1)^2 \equiv -3 \pmod p $$So $-3$ is quadratic residue. But this impossible since: $$ (\frac{-3}{p}) = (\frac{-1}{p})(\frac{3}{p}) = (-1)^{\frac{p-1}{2}} \cdot (-1)^{\frac{p-1}{2} \cdot \frac{3-1}{2}} \cdot (\frac{p}{3}) =(\frac{p}{3}) = (\frac{2}{3}) = -1 $$which is contradiction. Thus we are done.

I have a solution similar to dear Kimchiks926's which mainly using the Fermat's little theorem and the order in number theory. Assume that $p^2+p+1= x^3$,then we have $x^3\equiv1\pmod p$ ,and it's obvious that $(x,p)=1$, so we use the Fermat's little theorem and get $x^{p-1}\equiv 1\pmod p$, Assume that the order of x mod p is r,so we have $x^{r}\equiv 1\pmod p$, According to the quality of order,we have $r\mid (3,p-1)$,which leads to $r=1$ or $3$ if $r=1$,we have $x\equiv 1\pmod p$,assume that $x=kp+1(k\ge1)$,obviously we have $x^3=(kp+1)^3>p^2+p+1$,which is a contradiction. if $r=3$,we have $3\mid p-1$,assume that $p=3t+1(t\ge 1)$,we have $p^2+p+1=9t^2+9t+3=x^3$, then we$\mod 9$ on both sides,and we have $x^3\equiv 3\pmod 9$,which is impossible because $x^3$ can only $\equiv 0,-1 ,1\pmod 9$. Thus $p^2+p+1$ is never a perfect cube.

As $x^3\equiv \{1,-1, 0\}\mod 9$ and as $p^2+p+1$ must be a perfect cube, So $p\equiv -1\mod 9$ is only possible. So let $p=9k+8$ here $k$ is odd. Now let $p^2+p+1=x^3\implies p(p+1)=(x-1)(x^2+x+1)$ and we have $gcd(x,p)=1$ so $x^{p-1}\equiv x^{9k+7}\equiv 1\mod p$ and as $x^3\equiv x^{3*(3k+3)}\equiv x^{9k+7}*x^2 \equiv 1\mod p$ hence $x^2\equiv 1\mod p$ and as $x^2*x\equiv 1\mod p\implies x\equiv 1\mod p$ hence $x^2+x+1\equiv 3\mod p$ Now case 1-: $p|x^2+x+1$ which is only possible if $p=3$ and by checking at $p=3$ we can't get $p^2+p+1$ a perfect cube, Hence contradiction. Case 2 -: $p|x-1$ then let $x=pk_1+1$ then we have $k_1[(pk_1+1)^2+(pk_1+1)+1]=p+1$ which is clearly not possible, Hence contradiction.$\blacksquare$

I feel like I've seen this somewhere... Solution. Suppose there exists a prime $p$ and an integer $n$ such that $p^2 + p + 1 = n^3$. Rearranging and factoring gives \[p(p + 1) = (n - 1)(n^2 + n + 1),\]so either $p \mid n - 1$ or $p \mid n^2 + n + 1$. But if the first case holds then we have $n - 1\geq p$ and $n^2 + n + 1 \leq p + 1$, a contradiction. Thus $p \mid n^2 + n + 1 \mid n^3 - 1$ and $p \nmid n - 1$, so $\text{ord}_p(n) = 3$. This implies $p \equiv 1 \pmod 3$, and so $3 \mid n$. However, $p^2 + p + 1 \equiv 3 \pmod 9$, so $\nu_3(n^3) = 1$, a contradiction since this is at least $3$. $\square$