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

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Question

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

Sum
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Solution

Using binomial theorem, the given expression (3x* -2ax +3a* ) can be expanded

[3x2 - a (2x - 3a)]3

= 3C0 (3x2 - 2ax)3 + 3C1(3x2 - 2ax)2 (3a2)+ 3C2(3x2 - 2ax) (3a2)2 + 3C3(3a2)3

= (3x2 - 2ax)3 + 3(9x4 - 12ax3 + 4a2x2)(3a2)+3(3x2 - 2ax)(9a4) + 27a6

= (3x2 - 2ax)3 + 81a2x4 - 108a3x3 + 36a4x2 + 81a4x2 - 54a5x + 27a6

= (3x2 - 2ax)3 + 81a2x4 - 108a3x3 + 117a4x2 - 54a5x + 27a6

Again, by using binomial theorem, we obtain

(3x2 - 2ax)3

= 3C0 (3X2)3 - 3C1 (3X2)2 (2ax) + 3C2 (3X2)(2ax)2 - 3C3 (2ax)3

= 27x6 - 3(9x4) (2ax) + 3 (3x2) (4a2x2) -8a3x3

= 27x6 - 54ax5 + 36a2x4 - 8a3x3

From (1) and (2), we obtain

(3x2 - 2ax + 3a2)3

= 27x6 - 54ax5 + 36a2 x4 - 8a3x3 + 81a2x4 - 108a3x3  + 117a4 x2 - 54a5x + 27a6

= 27x6 - 54ax5 + 117a2 x4 - 116a3 x3 + 117a4 x2 - 54a5x + 27a6

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Chapter 7: Binomial Theorem - Miscellaneous Exercise [Page 131]

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NCERT Mathematics [English] Class 11
Chapter 7 Binomial Theorem
Miscellaneous Exercise | Q 6. | Page 131

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