<s>[INST] Primes such that $2p^2-3p-1$ is a perfect cube Find all primes $p$ such that $2p^2-3p-1$ is a perfect cube.
I've noticed that $p=2,3$ are solutions. Setting $2p^2-3p-1=x^3$ we can factorise this as $(4p-3)^2=8x^3+17$. However im unable to proceed further.
 [/INST] Let $$2p^2-3p-1=x^3$$
$$p(2p-3)=(x+1)(x^2-x+1)$$
If $p=2$ then $x=1$
If $p=3$ then $2p^2-3p-1=8=2^3$
Let $p>3$. Then $\gcd(p,2p-3)=1$ and $p<2p-3$
Then 
$$p=x+1$$
$$2p-3=x^2-x+1$$
Thus $$-3=x^2-3x-1$$
$$x^2-3x+2=0$$
Hence $x=1$ or $x=2$
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