Factorise and Expand

Given that:

$\begin{aligned} x^5 - x^4 & = x^4 - x & \small \blue{\text{Since }x \ne 0 \text{, we can divide}} \\ x^4 - x^3 & = x^3 - 1 & \small \blue{\text{both sides by }x.} \\ x^3(x-1) & = (x-1)(x^2 + x + 1) & \small \blue{\text{Since }x \ne 1 \text{, we can divide}} \\ x^3 & = x^2 + x + 1 & \small \blue{\text{both sides by }x-1.} \end{aligned}$

$\implies \red{x^3 - x^2 - x = 1}$

Now we have $(x+1)(x-1)^2 = (x^2-1)(x-1) = \red{x^3 - x^2 - x} + 1 = \red 1 + 1 = \boxed 2$ .

$x^5-x^4=x^4-x\implies x^3-x^2-x=1\implies x^3-x^2-x+1=2\implies (x+1)(x^2-2x+1)=2\implies (x+1)(x-1)^2=\boxed 2$