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Example 23 - Chapter 2 Class 9 Polynomials - Part 2

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.