Example 18 (ii) - Chapter 2 Class 9 Polynomials

Last updated at Dec. 16, 2024 by

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Example 23 Evaluate each of the following using suitable identities: (ii) (999) 3 We write 999 = 1000 – 1 (999) 3 = (1000 – 1) 3 Using (a – b)3 = a3 – b3 – 3ab(a – b) Where a = 1000 & b = 1 = (1000) 3 – (1) 3 – 3(1000)(1)(1000 – 1) = 1000000000 – 1 – 3(1000)(1)(999) = 1000000000 – 1 – 2997000 = 997002999

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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