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Ex 8.1 6 - Using Binomial Theorem, evaluate (96)3 - Chapter 8

Ex 8.1 6 - Chapter 8 Class 11 Binomial Theorem - Part 2

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Ex 8.1, 6 Using Binomial Theorem, evaluate (96)3 (96)3 = (100 – 4)3 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 b0 + 3C1 a2b1 + 3C2 a1 b2 + 3C3 a0 b3 = (3 !)/0!(3 − 0)! a3 × 1 + 3!/(1! (3 − 1) !) a2 b + 3!/(2! (3 −1)!) ab2 + 3!/(3 !(3 −3)!) 1 × b3 = 3!/3! a3 + (3 × 2!)/( 2!) a2 b + (3 × 2!)/2! ab2 + 3!/3! a3 = a3 + 3a2b + 3b2a + b3 Hence, (a + b)3 = a3 + 3a2 b + 3 ab2 + b3 Putting a = 100 & b = –4 (100 – 4)3 = (100)3 + 3(100)2 (–4) + 3(100) (–4)2 + (–4)3 (96)3 = 1000000 + 3(10000) (–4) + 3(100) (16) – 64 = 1000000 – 120000 + 4800 – 64 = 884736 Thus (96)3 = 884736

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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.