How can I factor $x^6+5x^3+8$? [duplicate]

$$x^6 + 5 x^3 + 8$$ I have tried to solve this by some middle term fact, but have failed and I think these are useless in this sum. Please help me. (Not a P.S.Q.)

This is a truly fiendish problem, well suited to Hallowe’en tide.

Set $t=x^3$, to get $t^2+5t+8=0$, whose roots are $(-5\pm\sqrt{-7})/2$. Now, the integers of $\Bbb Q(\sqrt{-7}\,)$ are a unique factorization domain, in which $$2=\frac{1+\sqrt{-7}}2\cdot\frac{1-\sqrt{-7}}2\,.$$ The number $(-5+\sqrt{-7})/2$ has Norm $8$, of course, that’s the constant term in its minimal polynomial. So you can hope that $t=(-5+\sqrt{-7})/2$ actually is a cube. Indeed, it’s equal to $\bigl[(1-\sqrt{-7}\,)/2\bigr]^3$. So we find that $(1+\sqrt{-7})/2$ is a root, and so $x^2-x+2$ is a factor of the original.

Alt. hint: $\,x=0\,$ is not a root, so dividing by $\,x \ne 0\,$ gives $\displaystyle\,x^3+\frac{8}{x^3}=-5\,$.

Let $\displaystyle\, z = x + \frac{2}{x}\,$, then $\displaystyle\,z^3 = x^3 + \frac{8}{x^3}+6z\,=6z -5 \;\implies\, z^3 - 6z + 5 = 0$. The latter has the obvious root $\,z=1\,$, so the original polynomial has a factor $\displaystyle\,x \cdot \left(x + \frac{2}{x}-1\right)=x^2-x+2\,$.