Question

facorization: x^6+5*x^3+8

Answer 1

let x^3 = y

   y^2  + 5 y + 8
  let us find roots of    y^2 + 5 y + 8 = 0
  discriminant :     squareroot(25 - 40) ]
  as the discriminant is negative, there are no real roots. 
  The given expression canno t be  factorized with real  factors.

     y1 = [- 5 + √15 i ] / 2              y2 = [ -5 - √15 ] / 2
 are the solutions which involve imaginary numbers.

      x⁶ + 5 x³ + 8  =     (2 x³ + 5 - √15 i) (2x³ + 5 - √15 i) / 4

Answer 2

This polynomial is specially made for using the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

With the idea of using above identity, split the middle term $5*x^3$ as $(-x)^3+6x^3=(-x)^3-3(x^2)(-x)(2)$ :

$x^6+5x^3+8=(x^2)^3+(-x)^3+2^3-3(x^2)(-x)(2)\\=(x^2-x+2)(x^4+x^2+4+x^3+2x-2x^2)\\=(x^2-x+2)(x^4+x^3-x^2+2x+4)$