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Write (a+b+c)12 as [a+(b+c)]12
(a+b+c)12 = [a+(b+c)]12 =
Then let d = (b+c)
(a+b+c)12 = [a+(b+c)]12 = (a+d)12
Use the binomial theorem to expand (a+d)12
There will be 13 terms, each term will contain a power of (b+c)
from the 12th power all the way down to the 0th power.
Use the binomial theorem to expand each of those powers of (b+c).
Incidentally there will be a total of 91 terms after all like terms
are collected.
Edwin