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