<s>[INST] Why does $(3\sqrt3)^2 = 27$? How does $(3\sqrt3)^2 = 27$?
I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.
 [/INST] If you really insist on expanding, just use the fact that $3\sqrt{3} = \sqrt{3} + \sqrt{3} + \sqrt{3}$, and therefore expand away $(\sqrt{3} + \sqrt{3} + \sqrt{3})^2$ using the trinomial expansion formula $(a+b+c)^2 = a^2 + b^2 +c^2 +2ab+2bc+2ac$.
Later edit:
And if you really insist on using FOIL, then use a little imagination to write $3\sqrt{3} = 2\sqrt{3} + \sqrt{3}$, thus your desired expression becoming $(2\sqrt{3} + \sqrt{3})(2\sqrt{3} + \sqrt{3})$.
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