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Question
Expand the following by using binomial theorem.
(2a – 3b)4
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Solution
(x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 +...+ nCr xn-r ar +...+ nCn an
∴ (2a - 3b)4 = 4C0 (2a)4 - 4C1 (2a)3 (3b)1 + 4C2 (2a)2 (3b)2 - 4C3 (2a)1 (3b)3 + 4C4 (3b)4
`= 1 xx 2^4 "a"^4 - (4) 2^3 "a"^3 (3"b") + (4 xx 3)/(2 xx 1) 2^2 "a"^2 3^2 "b"^2 - (4 xx 3 xx 2)/(3xx2xx1) (2"a")3^3 "b"^3 + 1(3^4"b"^4)`
= 16a4 - 96a3b + 216a2b2 - 216ab3 + 81b4
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