Help understanding how to factor completely $x^3-x^2-x+1$ I need someone to help explain the steps to completely factor the problem $x^3-x^2-x+1$.

Here is what I have done so far:


  $x^3-x^2-x+1$ 
  
  to 
  
  $x^3-x^2+-1(x+1)$
  
  Since there is a negative I changed $-x+1$ to $-1(x+1)$ 
  
  to 
  
  $x^2(x-1)+-1(x+1)$ 
  
  to
  
  $(x^2-1)(x-1)(x+1)$.


Is this correct? If not please explain what I did wrong and what I should do. If it is then thank you for saying so, if not, thank you for helping me to understand.
$$x^3 - x^2 - x + 1 \to x^2(x - 1) + (-1)(x-1)$$
$$\to (x-1)(x^2 - 1) \to (x-1)^2(x+1)$$
\begin{align}x^3-x^2-x+1&amp;=x^3-x^2-x+1+(3x^2-3x^2)+(3x-3x)\\
&amp;=x^3+3x^2+3x+1-x^2-x-3x^2-3x\\
&amp;=(x+1)^3-4x^2-4x\\
&amp;=(x+1)^3-4x(x+1)\\
&amp;=(x+1)\left[(x+1)^2-4x\right]\\
&amp;=(x+1)(x-1)^2\end{align}