How do you foil \[{\left( {x - 1} \right)^2}\] ?

Hint:
In this, there will be four-term which are first, inner, outer, and last. Here all the terms will get multiplied with their respective binomials, and when this gets done then we will simplify the expression and in this way, we will get the equation.

Complete step by step solution:
we have the expression given as \[{\left( {x - 1} \right)^2}\] , and we have to foil this expression.
So, the expression ${\left( {x - 1} \right)^2}$ can be represented as
$ \Rightarrow {x^2} - 2 \cdot x \cdot 1 + {1^2}$
And on solving the above equation it can be written as
\[ \Rightarrow {x^2} - x - x + 1\]
And taking the common and solving it we get
\[ \Rightarrow x\left( {x - 1} \right) - 1\left( {x - 1} \right)\]
And it will be equal to
$ \Rightarrow \left( {x - 1} \right)\left( {x - 1} \right)$
Now we will express this in the form of foil. So for this, we will first see the term we had used in the hint part.
So, the first term will be $x \cdot x$
The outside term will be equal to $x \cdot \left( { - 1} \right)$
The inside term will be equal to $\left( { - 1} \right) \cdot x$
And the last term will be equal to $\left( { - 1} \right) \cdot \left( { - 1} \right)$
So combining together all the term we will get the equation as
$ \Rightarrow x \cdot x + x \cdot \left( { - 1} \right) + \left( { - 1} \right) \cdot x + \left( { - 1} \right) \cdot \left( { - 1} \right)$
And solving the above equation we will get the equation as
$ \Rightarrow {x^2} - x - x + 1$
And solving the like terms, we get
$ \Rightarrow {x^2} - 2x + 1$

Therefore, by using the foil the equation will become $x \cdot x + x \cdot \left( { - 1} \right) + \left( { - 1} \right) \cdot x + \left( { - 1} \right) \cdot \left( { - 1} \right)$.

Note:
FOIL is a mnemonic and it helps to remember the combinations of terms that need to be combined and can be evaluated as the product of two binomials. For more understanding, we can remember that when we multiply two terms then we must have to multiply the coefficient and then add the exponents. So the same process we do in foil too.