Skorzystamy ze wzoru skróconego mnożenia na różnicę sześcianów.

$W\left(x\right)=x\left(2x^2-x\right)+x-2$

$W\left(x\right)=x^3+x^3-x^2+x-1-1$

$W\left(x\right)=\left(x^3-1\right)+\left(x^3-x^2+x-1\right)$

$W\left(x\right)=\left(x-1\right)\left(x^2+x+1\right)+\left[\left(x^3-1\right)-\left(x^2-x\right)\right]$

$W\left(x\right)=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(x^2+x+1\right)-x\left(x-1\right)$

$W\left(x\right)=\left(x-1\right)\left(x^2+x+1+x^2+x+1-x\right)$

Odpowiedź:

$W\left(x\right)=\left(x-1\right)\left(2x^2+x+2\right)$