Find the expansion (3x2−2ax+3a2)3 using binomial theorem.
For binomial expansion first let's do a small pairing inside the bracket
[3x2−a(2x−3a)]3
Now let's expand this as be normally do for two digit expansion.
[3x2−a(2x−3a)]3=(3x2)3−3C1(3x2)2.a(2x−3a)+3C2(3x2).a2(2x−3a)2−a3(2x−3a)327x6−54x5a+81a2x4−108a3x3+81a4x3−8a3x3+36a4x2−54a5x+27a6
27x6−54x5a+117a2x4−116a3x3+117a4x3−54a5x+27a6