Q10.    Find the expansion of (3x^2 - 2ax +3a^2)^3 using binomial theorem.

Answers (1)

Given (3x^2 - 2ax +3a^2)^3

By Binomial Theorem It can also be written as 

(3x^2 - 2ax +3a^2)^3=((3x^2 - 2ax) +3a^2)^3

= ^3C_0(3x^2-2ax)^3+^3C_1(3x^2-2ax)^2(3a^2)+^3C_2(3x^2-2ax)(3a^2)^2+^3C_3(3a^2)^3

= (3x^2-2ax)^3+3(3x^2-2ax)^2(3a^2)+3(3x^2-2ax)(3a^2)^2+(3a^2)^3

= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+36a^4x^2+81a^4x^2-54a^5x+27a^6

= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6...........(1)

Now, Again By Binomial Theorem,

(3x^2-2ax)^3=^3C_0(3x^2)^3-^3C_1(3x^2)^2(2ax)+^3C_2(3x^2)(2ax)^2-^3C_3(2ax)^3

(3x^2-2ax)^3= 27x^6-3(9x^4)(2ax)+3(3x^2)(4a^2x^2)-8a^2x^3

(3x^2-2ax)^3= 27x^6-54x^5+36a^2x^4-8a^3x^3............(2)

From (1) and (2) we get,

(3x^2-2ax+3a^2)^3=27x^6-54x^5+36a^2x^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6

(3x^2-2ax+3a^2)^3=27x^6-54x^5+117a^2x^3-116a^3x^3+117a^4x^2-54a^5x+27a^6

 

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