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

Hint: Factorization or the process of finding factor is the method where we find the zeroes of root of a function. For example, we have $f\left( x \right)$, we will have to find those values of $x$ for which the entire function $f\left( x \right)$would be equal to $0$.But since, our expression is not equated to $0$, we don’t have to find it’s roots. We just have to express our expression in terms of it’s factors. For a simple quadratic expression, we have a formula to find it’s factor but as the order of the polynomial increases, we’d either have to find it’s factors by hit or trial method or by grouping.

Complete step by step solution:
Let us find the factors of the expression given in the question through grouping as asked in the question.
Now let us consider $f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1$.
Let us first group some terms together so we can see what formulae we can apply.
Let us group the ${{x}^{3}}$ term and the constant term as we can apply the ${{a}^{3}}-{{b}^{3}}$formula.
Upon grouping, we get the following :
$\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-1+{{x}^{2}}-x \\
\end{align}$
The other two terms got grouped automatically.
We all know the formula of ${{a}^{3}}-{{b}^{3}}$.
It is ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.
Upon comparing, our $a=x,b=1$.
Upon applying the formula, we get the following :
\[\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-1+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+{{x}^{2}}-x \\
\end{align}\]
Now let us take an $x$ common from the term ${{x}^{2}}-x$.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-1+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+x\left( x-1 \right) \\
\end{align}\]
Now let us take $\left( x-1 \right)$common from the above equation.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-1+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+x\left( x-1 \right) \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left[ {{x}^{2}}+1+x+x \right] \\
\end{align}\]
Now, let us add both the $x$.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-1+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+x\left( x-1 \right) \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left[ {{x}^{2}}+1+x+x \right] \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left[ {{x}^{2}}+2x+1 \right] \\
\end{align}\]
We all know that ${{\left( x+1 \right)}^{2}}={{x}^{2}}+2x+1$.
Let us plug it in the above equation.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-1 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-1+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+{{x}^{2}}-x \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+1+x \right)+x\left( x-1 \right) \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left[ {{x}^{2}}+1+x+x \right] \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right)\left[ {{x}^{2}}+2x+1 \right] \\
 & \Rightarrow f\left( x \right)=\left( x-1 \right){{\left( x+1 \right)}^{2}} \\
\end{align}\]
$\therefore $ Hence, the factors of the expression ${{x}^{3}}+{{x}^{2}}-x-1$ that we got by grouping are ${{x}^{3}}+{{x}^{2}}-x-1=\left( x-1 \right){{\left( x+1 \right)}^{2}}$.

Note: We are not finding the roots of this particular expression. Although $\pm 1$ are the roots of this equation, we just have left this expression as a product of its factors since it’s not equated to $0$.
We have to be careful while taking common terms. We should remember all the algebra formulae to complete the question quickly. We can also do it through trial and error where we try guessing one of its factors and substitute it in the given equation.