Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.
We know that-
(a + b)3 = a3 + 3a2b + 3ab2 + b3
putting a = 3x2 & b = -a(2x-3a), we get-
[3x2 + (-a(2x-3a))]3
= (3x2)3+3(3x2)2(-a(2x-3a)) + 3(3x2)(-a(2x-3a))2 + (-a(2x-3a))3
= 27x6 - 27ax4(2x-3a) + 9a2x2(2x-3a)2 - a3(2x-3a)3
= 27x6 - 54ax5 + 81a2x4 + 9a2x2(4x2-12ax+9a2)
- a3[(2x)3 - (3a)3 - 3(2x)2(3a) + 3(2x)(3a)2]
= 27x6 - 54ax5 + 81a2x4 + 36a2x4 - 108a3x3 + 81a4x2
-8a3x3 + 27a6 + 36a4x2 - 54a5x
= 27x6 - 54ax5+ 117a2x4 - 116a3x3 + 117a4x2 - 54a5x + 27a6
Thus, (3x2 – 2ax + 3a2)3
= 27x6 - 54ax5+ 117a2x4 - 116a3x3 + 117a4x2 - 54a5x + 27a6