Example 23 - Chapter 2 Class 9 Polynomials - Part 2

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  1. Chapter 2 Class 9 Polynomials (Term 2)
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.