How do you FOIL where three factors are used \[\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\]?

Hint: Here, we will first multiply last two factors using foil method and distributive property. Then we will multiply the first factor to the expression obtained by multiplying to the last two factors. We will then add and subtract the like terms to get the required expression. The given factors are linear expression and as there are three factors so when these factors will be multiplied we will get a cubic expression.

Complete step-by-step answer:
We know that the FOIL method can be represented by the equation \[\left( {a + b} \right)\left( {c + d} \right) = a \cdot c + a \cdot d + b \cdot c + b \cdot d\].
When three factors are used, it is not possible to simplify the expression just by using the FOIL method.
The FOIL method and the distributive law of multiplication can be used together to simplify a product of three factors.
The given expression is \[\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\].
First, we will multiply the second and third parentheses using the FOIL method.
Therefore, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left[ {x \cdot x + x\left( { - 3} \right) + 2 \cdot x + 2\left( { - 3} \right)} \right]\]
Multiplying the terms in the parentheses, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left( {{x^2} - 3x + 2x - 6} \right)\]
Adding the like terms in the expression, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left( {{x^2} - x - 6} \right)\]
Now, the right hand side is the product of a binomial and a trinomial.
We will group two terms in the trinomial.
Rewriting the expression, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left[ {{x^2} - \left( {x + 6} \right)} \right]\]
Multiplying the two parentheses with the help of FOIL method, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = x \cdot {x^2} + x\left[ { - \left( {x + 6} \right)} \right] - 1 \cdot {x^2} - 1\left[ { - \left( {x + 6} \right)} \right]\]
Simplifying the terms in the expression, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = {x^3} - x\left( {x + 6} \right) - {x^2} + 1\left( {x + 6} \right)\]
Now, we can simplify the remaining expression using the distributive law of multiplication.
Multiplying the terms using the distributive law of multiplication \[a\left( {b + c} \right) = a \cdot b + a \cdot c\], we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = {x^3} - x \cdot x - x \cdot 6 - {x^2} + 1 \cdot x + 1 \cdot 6\]
Multiplying the terms in the expression, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = {x^3} - {x^2} - 6x - {x^2} + x + 6\]
Combining the like terms in the expression, we get
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right) = {x^3} - 2{x^2} - 5x + 6\]
Therefore, we get the value of the product \[\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\] as \[{x^3} - 2{x^2} - 5x + 6\].

Note: We have used the distributive law of multiplication in the solution to multiply a monomial by a binomial in the solution. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
The FOIL method stands for First, Outer, Inner, Last.
First means multiplication of the first term of the first parentheses and the first term of the second parentheses.
Outer means multiplication of the first term of the first parentheses and the second term of the second parentheses.
Inner means multiplication of the second term of the first parentheses and the first term of the second parentheses.
Last means multiplication of the second term of the first parentheses and the second term of the second parentheses.