Polynomial Diophantine Equation 1

Given : $x^6+3x^4+6x^2+5=y^3$

Consider the $\text{L.H.S.}$

$x^6+3x^4+6x^2+5$

$= x^6+3x^4+3x^2 + 1 + 3x^2 +3 +1$

$= (x^2+1)^3 +3(x^2+1) + 1$

Let $t=x^2+1 \ ; \ t>0$

$\implies t^3 + 3t +1 =y^3$

Note that $t^3 < t^3+3t+1 < (t+1)^3$

But an integral cube cannot exist in between two consecutive integral cubes.

$\implies \text{Number of solutions} = \boxed{0}$