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Question

Mathematics Question on Binomial theorem

Find the expansion of (3x2–2ax+3a2)3(3x^2 – 2ax + 3a^2)^3 using binomial theorem.

Answer

Using Binomial Theorem, the given expression (3x2−2ax+3a2)3(3x^2-2ax +3a^2)^3 can be expanded as
[(3x2−2ax)+3a2]3[(3x^2-2ax)+3a^2]^3
= 3C0(3x2−2ax)3+  3C1(3x2−2ax)2(3a2)+  3C2(3x2−2ax)(3a2)2+  3C3(3a2)2^3C_0 (3x^2 -2ax)^3 + \space^3C_1(3x^2 - 2ax)^2 (3a^2) +\space ^3C_2 (3x^2 - 2ax) (3a^2)^2 + \space^3C_3 (3a^2)^2
=(3x2−2ax)3+3(9x4−12ax3+4a2x2)(3a2)+3(3x2−2ax)(9a4)+27a6=(3x^2-2ax)^3+3(9x^4-12ax^3 +4a^2x^2) (3a^2)+3(3x^2-2ax) (9a^4)+27a^6
=(3x2−2ax)3+81a2x4−108a3x3+36a4x2+81a4x2−54a5x+27a6=(3x^2-2ax)^3 +81a^2x^4-108a^3x^3 +36a^4x^2+81a^4x^2-54a^5x+27a^6
=(3x2−2ax)3+81a2x4−108a3x3+117a4x2−54a5x+27a6...(1)=(3x^2-2ax)^3 +81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6 ...(1)

Again by using Binomial Theorem, we obtain
(3x2−2ax)3(3x^2-2ax)^3
=  3C0(3x2)3−  3C1(3x2)2(2ax)+  3C2(3x2)(2ax)2−3C3(2ax)3=\space^ 3C_0 (3x^2)^3 - \space^3C_1 (3x^2)^2 (2ax) + \space^3C_2 (3x^2) (2ax)^2 - 3C^3 (2ax)^3
=27x6−3(9x4)(2ax)+3(3x2)(4a2x2)−8a3x3=27x^6-3(9x^4) (2ax)+3(3x^2) (4a^2x^2)-8a^3x^3
=27x6−54ax5+36a2x4+−8a3x3...(2)=27x^6-54ax^5 +36a^2x^4+-8a^3x^3 ...(2)

From (1) and (2), we obtain
(3x2−2ax+3a2)3(3x^2-2ax +3a^2)^3
=27x6−54ax5+36a2x4−8a3x3+81a2x4−108a3x3+117a4x2−54a5x+27a6=27x^6-54ax^5 +36a^2x^4-8a^3x^3 +81a^2x^4-108a^3x^3 +117a^4x^2 - 54a^5x+27a^6
=27x6−54ax5+117a2x4+−116a3x3+117a4x2−54a5+27a6=27x^6-54ax^5+117a^2x^4+-116a^3x^3 +117a^4x^2 -54a^5 +27a^6