How to solve the diophantine equation $x^2-x=y^3$

Question

How would I solve the diophantine equation $x^2-x=y^3$?

Would anyone be able to help solve this equation?

We have only done equations up to powers of $2$ as of now.

Answer 1

$y^3=x^2-x=x(x-1)$ In the prime factorization of $y^3,$ the exponent of every prime is a multiple of $3.$ Since $x, x-1$ are coprime, none of these primes occurs in both the factorization of $x$ and of $x-1$ so both $x$ and $x-1$ are cubes.

I'm sure you can take it from here.