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Question
Expand: `(2x - 1/x)^6`
Solution
`(2x - 1/x)^6 = ""^6"C"_0(2x)^6 - ""^6"C"_1(2x)^5(1/x) + ""^6"C"_2(2x)^4(1/x)^2 - ""^6"C"_3(2x)^3(1/x)^3 + ""^6"C"_4(2x)^2(1/x)^4 - ""^6"C"_5(2x)(1/x)^5 + ""^6"C"_6(1/x)^6`
Now, 6C0 = 1 = 6C6
6C1 = 6 = 6C4
6C2 = `(6 xx 5)/(1 xx 2)` = 15 = 6C4
6C3 = `(6 xx 5 xx 4)/(1 xx 2 xx 3)` = 20
∴ `(2x - 1/x)^6 = 1 xx 64x^6 - 6 xx 32x^5 xx 1/ x + 15 xx 16x^4 xx 1/x^2 - 20 xx 8x^3 xx 1/ x^3 + 15 xx 4x^2 xx 1/ x^4 - 6 xx 2x xx 1/x^5 + 1 xx 1/x^6`
= `64x^6 - 192x^4 + 240x^2 - 160 + 60/x^2 - 12/x^4 + 1/x^6`.
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