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प्रश्न
Answer the following:
Expand `((2x)/3 - 3/(2x))^4`
उत्तर
`((2x)/3 - 3/(2x))^4 = ""^4"C"_0 ((2x)/3)^4 - ""^4"C"_1 ((2x)/3)^3 (3/(2x)) + ""^4"C"_2 ((2x)/3)^2 (3/(2x))^2 - ""^4"C"_3 ((2x)/3)(3/(2x))^3 + ""^4"C"_4 (3/(2x))^4`
Now, 4C0 = 1 = 4C4
4C1 = 4 = 4C3
4C2 = `(4 xx 3)/(1 xx 2)` = 6
∴ `((2x)/3 - 3/(2x))^4 = 1 xx (16x^4)/81 - 4 xx (8x^3)/27 xx 3/(2x) + 6 xx (4x^2)/9 xx 9/(4x^2) - 4 xx (2x)/3 xx 27/(8x^3) + 1 xx 81/(16x^4)`
= `(16x^4)/81 - (16x^2)/9 + 6 - 9/x^2 + 81/(16x^4)`
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