Ex 2.4, 7 (i) - Chapter 2 Class 9 Polynomials (Important Question)

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Ex 2.4, 7 Evaluate the following using suitable identities: (i) (99)3 We write 99 = 100 – 1 (99)3 = (100 − 1)3 Using (a – b)3 = a3 – b3 – 3ab(a – b) Where a = 100 & b = 1 = (100)3 − (1)3 − 3(100) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo