2015/040) Factorize $x^6+5x^3+8$

we checking rational solutions find that there is no rational root and hence there is no term of the form (x+a)

now $x^6 = (x^2)^3$ and $8= 2^3$ if we can express $5x^3$ so that one is a cube and second one is product of $x^2$,$-x$ and the 3rd term then we get into the form $a^3+b^3+c^3-3abc$

$x^6+5x^3+8 = x^6-x^3 + 2^3 -6x^3$
= $x^6 - x^3 + 2^3 - 3(x^2)(x)$
= $a^3+b^3+c^3 - 3abc$ where $a = x^2,b = -x,c = +2$

hence it factors as $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
 = $(x^2 -x +2)(x^4+x^2+4 + x^3 -2x - 2x^2)$
= $(x^2-x+2)(x^4+x^3-x^2+2x+4)$