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Misc 10 - Find expansion of (3x2 - 2ax + 3a2)3 using binomial

Misc 10 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc 10 - Chapter 8 Class 11 Binomial Theorem - Part 3


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Misc 10 Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem. We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)3 = 3C0 a3 + 3C1 a2b1 + 3C2 a1 b2 + 3C3 b3 = a3 + 3!/(1! (3 −1) !) a2 b + 3!/2!(3 −1)! ab2 + b3 = a3 + 3a2b + 3b2a + b3 Now, (3x2 – 2ax + 3a2)3 = (3x2 – (2ax – 3a2))3 = (3x2 – a(2x – 3a))3 Putting a = 3x2 and b = –a( 2x – 3a) in (1) (a + b)3 = a3 + 3a2b + 3b2a + b3 (3x2 – a (2x – 3a2))3 = (3x2)3 + 3(3x2)2 (–a(2x – 3a)) + 3 (–a(2x – 3a))2 (3x2) + (–a(2x – 3a))3 = 27x6 – 27x4a (2x – 3a) + 9x2a2 (2x – 3a)2 – a3 (2x – 3a)3 = 27x6 – 27x4 (2xa – 3a2) + 9x2a2 (4x2 + 9a2 − 12ax) – a3 ((2x)3 + (−3a)3 + 3(2x)2 (−3a) + 3 (2x) (−3a)2] = 27x6 – 54x5a + 81 x4a2 + 36x4a2 + 81x2a4 – 108x3a3 – a3 (8x3 – 27a3 – 36x2a + 54xa2) = 27x6 – 54x5a + 81 x4a2 + 36x4a2 + 81x2a4 – 108x3a3 – 8x3 a3 + 27a6 + 36x2a4 – 54a5x = 27x6 – 54ax5 + 81a2x4 + 36a2x4 – 108x3a3 – 8x3 a3 + 36x2a4 + 81x2a4 – 54a5 x + 27a6 = 27x6 – 54ax5 + 117a2x4 – 116a3x3 + 117x2a4 – 54a5 x + 27a6 Thus, (3x2 – 2ax + 3a2)3 = 27x6 – 54ax5 + 117a4x2 – 116a3x3 – 54a5 x + 27a6

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.