How to solve $y^3=x(x+1)$ where $x$ and $y$ are integers?

Question

How to solve $y^3=x(x+1)$ where $x$ and $y$ are integers ? Can you help me ?

Thanks :)

Answer 1

Hint: notice that $gcd(x,x+1)=1$ and thus $x=a^3$ and $x+1=b^3$

Answer 2

Hint

If two numbers $u$, $v$ are coprime and $uv$ is a $r$-th perfect power, then both $u$ and $v$ are $r$-th perfect powers.