Ex 2.5,7 - Chapter 2 Class 9 Polynomials - Part 3

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

Transcript

Ex 2.5, 7 Evaluate the following using suitable identities: (iii) (998)3 We write 998 = 1000 – 2 (998)3 = (1000 − 2)3 Using (a – b)3 = a3 – b3 – 3ab(a – b) Where a = 1000 & b = 2 = (1000)3 − (2)3 − 3(1000) (2) (1000 − 2) = 1000000000 − 8 − 6000(998) = 1000000000 − 8 − 5988000 = 1000000000 − 5988008 = 994011992

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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.