Difference of consecutive cubes never divisible by 5.

$$x^2+x\equiv 3\pmod 5\\ \implies x^2+x+4^{-1}\equiv 3+4^{-1}\pmod 5\\ \implies (x+2^{-1})^2=(x+3)^2\equiv 2\pmod 5$$

The quadratic residues modulo $5$ are $0,1,4$ and this list does not include $2$. QED

Note that by Little Fermat, $$ (x^3)^3\equiv x^{2\phi(5)+1}\equiv x\pmod{5} $$ Thus, if $x^3\equiv y^3\pmod{5}$, by cubing both sides, we must have $x\equiv y\pmod{5}$. Therefore, $$ 5\mid x^3-y^3\implies5\mid x-y $$

$x^3 - y^3 = (x - y)(x^2 + x y + y^2)$. Now $x^2 + x y + y^2 \equiv (x + 3 y)^2 - 3 y^2 \mod 5$, and since $3$ is not a quadratic residue mod $5$ we find that there are no solutions to $x^2 + x y + y^2 \equiv 0 \mod 5$ other than the trivial $(0,0)$. Thus
$x^3 \equiv y^3 \mod 5$ only when $x \equiv y \mod 5$.