How do you factor ${{x}^{4}}-2{{x}^{3}}+2x-1$?

Hint: First we explain the process of grouping. We then apply the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to complete the factorisation. We take the common divisors out to form one more common term. We have to use the identity one more time to complete the factorisation. We verify the factorisation using the vanishing method.

Complete step by step answer:
Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.
In case of ${{x}^{4}}-2{{x}^{3}}+2x-1$, the grouping will be done for ${{x}^{4}}-1$ and $-2{{x}^{3}}+2x$.
We apply the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for ${{x}^{4}}-1$. The replacements are $a={{x}^{2}},b=1$. The equation becomes ${{x}^{4}}-1={{\left( {{x}^{2}} \right)}^{2}}-{{1}^{2}}$. Applying the identity, we get
${{x}^{4}}-1={{\left( {{x}^{2}} \right)}^{2}}-{{1}^{2}}=\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)$.
For $-2{{x}^{3}}+2x$, we try to take the common numbers out.
For $-2{{x}^{3}}+2x$, we take $-2x$ and get $-2x\left( {{x}^{2}}-1 \right)$.
The equation becomes
$\begin{align}
  & {{x}^{4}}-2{{x}^{3}}+2x-1 \\
 & =\left( {{x}^{4}}-1 \right)+\left( -2{{x}^{3}}+2x \right) \\
 & =\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)-2x\left( {{x}^{2}}-1 \right) \\
\end{align}$.
Both the terms have $\left( {{x}^{2}}-1 \right)$ in common. We take that term again and get
$\begin{align}
  & \left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)-2x\left( {{x}^{2}}-1 \right) \\
 & =\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1-2x \right) \\
 & =\left( {{x}^{2}}-1 \right){{\left( x-1 \right)}^{2}} \\
\end{align}$
Although the grouping is completed, the factorisation is still remaining.
Now we find the factorisation of the equation ${{x}^{2}}-1$ using the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We convert to the equation of ${{x}^{2}}-1={{x}^{2}}-{{1}^{2}}$.
Therefore, we get ${{x}^{2}}-{{1}^{2}}=\left( x+1 \right)\left( x-1 \right)$. The individual terms are in their primitive form. We can’t further break those polynomials.
The factorisation is ${{x}^{4}}-2{{x}^{3}}+2x-1=\left( {{x}^{2}}-1 \right){{\left( x-1 \right)}^{2}}=\left( x+1 \right)\left( x-1 \right)\left( x-1 \right)\left( x-1 \right)$.

Note: We find the value of x for which the function $f\left( x \right)={{x}^{4}}-2{{x}^{3}}+2x-1$. We can see $f\left( 1 \right)={{1}^{4}}-2{{\left( 1 \right)}^{3}}+2\times 1-1=1-2+2-1=0$. So, the root of the $f\left( x \right)={{x}^{4}}-2{{x}^{3}}+2x-1$ will be the function $\left( x-1 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Now, $f\left( x \right)={{x}^{4}}-2{{x}^{3}}+2x-1=\left( x+1 \right)\left( x-1 \right)\left( x-1 \right)\left( x-1 \right)$. We can also do the same process for $\left( x+1 \right)$.