# Is there a formula for the binomial expansion of $(a-b)^n$?

Like there is a formula for the binomial expansion of $(a+b)^n$ that can be neatly and compactly be written as a summation, does there exist an equivalent formula for $(a-b)^n$ ?

• Try letting $x = -b$. Jul 16, 2016 at 11:11
• Yes, just substitute for b for -b Jul 16, 2016 at 12:27
• Ok...so you know the formula for $(a+b)^n$...now,try this...$(a-b)^n=\{a+(-b)\}^n$....Now,you know the formula..... Jul 16, 2016 at 12:54

Hint:$$\\ (a+b)^{ n }=\sum _{ k=0 }^{ n }{ \binom{n}{k} { a }^{ n-k } } { b }^{ k }$$ now substitute here $$b=-b$$ $$(a-b)^{ n }=\sum _{ k=0 }^{ n }{ { \left( -1 \right) }^{ k }\binom{n}{k} { a }^{ n-k } } { b }^{ k }$$
• This answer could be made clearer. By setting $b=-b$ you are effectively setting $b=0$. While I realise your intent, it could lead to confusion for others. Jul 16, 2016 at 13:05
• maybe $b\mapsto -b$ would be more clear notation Dec 10, 2020 at 14:16
$$(a-b)^n = (a+ (-b))^n = \sum_{k=0}^n \binom{n}{k}a^k (-b)^{n-k} = \sum_{k=0}^n \binom{n}{k}(-1)^{n-k}a^kb^{n-k} =$$