Is there any known way to multiply three binomials using a method similar to the FOIL method? I have searched the internet and have not found any such method.
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$\begingroup$You'll add the products of all possible combinations of terms, taking one term from each binomial.
This is the "generalized FOIL method."
For a product of two binomials, there are four combinations of two terms, taking one from each binomial: the First, Outer, Inner, and Last.
For a product of three binomials, there are eight combinations of three terms, taking one from each: all of the firsts (one combination), all of the lasts (one combination), one first and two lasts (three combinations), and two first and one last (three combinations).
For a product of nine binomials, you'll have 512 combinations of nine terms, taking one from each.
And so forth.
$\endgroup$ $\begingroup$As John answered; you're adding all the possible combinations of terms. A visualization of your standard FOIL operation with 2 binomials looks like such:
Here we are taking $a$ then multiplying by both $c$ and $d$. Then we do the same for $b$.
When we have three binomials however, it looks more like this:
Here we are taking $a$, then multiplying by both $c$ and $d$; but then we take the answers we got from each of those and for those two; each is multiplied by both $e$ and $f$. This is referring to John's answer about all the possible combinations. With 2 binomials FOIL gives you all the combinations. The arrows here show how to do this in one* step.
*such as FOIL is somewhat, one step
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