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.
 A: $$x^3 - x^2 - x + 1 \to x^2(x - 1) + (-1)(x-1)$$
$$\to (x-1)(x^2 - 1) \to (x-1)^2(x+1)$$
A: Let $f(x)=x^3-x^2-x+1$. By Rational Root Theorem, any rational root $\frac{p}{q}$ of $g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ has $p\mid a_0$ and $q\mid a_n$ (i.e. $q$ divides $a_n$). Proof is simple: substitute $x=\frac{p}{q}$, multiply both sides by $q^n$, observe divisibility.     
Since $f(1)=0$, we know $x^3-x^2-x+1=(x-1)P(x)$ for some $P(x)$, which of course must be quadratic, so $P(x)=x^2+ax+b$.   
$$(x-1)(x^2+ax+b)=x^3+(a-1) x^2+(b-a)x-b=x^3-x^2-x+1$$    
So $a-1=-1, b-a=-1, -b=1$, so $b=-1, a=0$.     
$$x^3-x^2-x+1=(x-1)(x^2-1)$$
A: Simply notice that $x=1$ is a root. Hence by Ruffini your polynomial $x^3-x^2-x+1$ is divisible by $x-1$. Thus there exists a second degree polynomial $p(x)$ (you can find by a simple polynomial division) such that $x^3-x^2-x+1=(x-1)p(x)$. Being $p$ of degree $2$ you can easily find its roots with the well known $\Delta$ formula, in order to factorize completely your polynomial.
A: we want to factor the polynomial $\mathbb{P}=x^{3}-x^{2}-x+1$ 
factor out $x^2$ to start as follows;
$\mathbb{P}=x^2(x-1)-(x-1)$
This makes sense because if you expand above you would get, $x^3-x^2-(x-1)=x^3-x^2-x+1$ which is of the form you had at the start.
Now factor out $(x-1)$
$$=(x-1)(x^2-1^2)$$
It may be helpful to see it as,
$(x-1)(x^2-1)$ say we call $(x-1)=a$ then you can see above on the second line for example we had $ax^2-a$. So to factor out a, can you see that it would be $a(x^2-1)$
I added the square because of the next step, and it does not change the one of course. 
Now, we know that $x^2-y^2=(x+y)(x-y)$ You can expand this to see that it is true if you like.
so $(x^2-1^2)=(x-1)(x+1)$
so $\mathbb{P}=(x-1)(x-1)(x+1)=(x-1)^2(x+1)$
Of course there are other ways that the polynomial can be factored, but this seems to be the nicest final form
A: \begin{align}x^3-x^2-x+1&=x^3-x^2-x+1+(3x^2-3x^2)+(3x-3x)\\
&=x^3+3x^2+3x+1-x^2-x-3x^2-3x\\
&=(x+1)^3-4x^2-4x\\
&=(x+1)^3-4x(x+1)\\
&=(x+1)\left[(x+1)^2-4x\right]\\
&=(x+1)(x-1)^2\end{align}
