## Precalculus (6th Edition) Blitzer

$64a^6+192a^5b+240a^4b^2+160a^3b^3+60a^2b^4+12ab^5+b^6$
Based on the Binomial Theorem, we have $(2a+b)^6=\binom60 (2a)^6(b)^0+\binom61 (2a)^5(b)^1+\binom62 (2a)^4(b)^2+\binom63 (2a)^3(b)^3+\binom64 (2a)^2(b)^4+\binom65 (2a)^1(b)^5+\binom66 (2a)^0(b)^5 =64a^6+192a^5b+240a^4b^2+160a^3b^3+60a^2b^4+12ab^5+b^6$