Show that $x^6 - x^4 +2 >0$ for all $x$ 
Show that $x^6 - x^4 +2 >0$ for all $x$.

I can prove this using calculus and I can prove it by factorising the expression. 
Is there any way to prove it another way, perhaps by considering cases of $x$? I need a simpler way to explain to a student that cannot factorise degree $6$ polynomials and use calculus at this stage. They can factorise degree $2$ though, if that helps!
Thanks.

Use sum of squares! $x^6>x^4$ for $|x|>1$
Otherwise $|x|<1$ and we have $x^6-x^4+2=(x^3-\frac{1}{2}x)^2-\frac{1}{4}x^2+2>0$ since $2>\frac{1}{4}x^2$
Or, magically, we have
$x^6-x^4+2=(x^3-x)^2+(x^2-1)^2+x^2+1>0$
