Question

Factorise $x^6+5x^3+8$

Answer 1

Given $x^6+5x^3+8$

Adding and subtracting $x^3$

$=x^6+5x^3+8+x^3-x^3$

$=x^6+6x^3+8-x^3$

$=(x^2{)}^3+(2{)}^3+(-x{)}^3-3(x^2\left)\right(2\left)\right(-x)$

According to the identity

$[x^3+y^3+z^3-3xyz=(x+y+z\left)\right(x^2+y^2+z^2-xy-yz-zx\left)\right]$

$=(x^2+2-x)\left[\right(x^2{)}^2+(2{)}^2+(-x{)}^2-\left(x^2\right)\left(2\right)-\left(2\right)(-x)-\left(x^2\right)(-x)]$

$=(x^2-x+2)(x^4+x^2+4-2x^2+2x+x^3)$

$=(x^2-x+2)(x^4+x^3-x^2+2x+4)$

Thus, factors of $x^6+5x^3+8$ are $(x^2-x+2)(x^4+x^3-x^2+2x+4)$