How many solutions?

Given: $6x^2+2=y^3$

Clearly, $y\ge2$ since the L.H.S. is always $>1$ . Further, we may also assume that $x>0$ since $x^2$ is an even function.

$\implies 2|y^3 \implies y=2a \: ; \: a\ge 1$

$\implies 3x^2+1=4a^3$

If $x$ is even $\implies$ L.H.S. is odd and we will have a contradiction.

$\implies x$ is odd.

Let $x=2b+1 \: ; \: b\ge 0$

$\implies 3b^2+3b+1=a^3$

$\implies (b+1)^3-b^3=a^3$

$\implies a^3+b^3=(b+1)^3$

By Fermat's Last Theorem , the above equation has no solution in positive integers . But, since $b\ge0$ , it can have solutions in whole numbers.

So, putting $b=0$ , we find that $a=1$ which gives $y=2$ and $x=1$ and since $x^2$ is an even function, $x=-1$ is also a solution.

So, the solutions are $(1,2) \: \text{and} \: (-1,2)$ .