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Ex 2.4, 7 (iii) - Chapter 2 Class 9 Polynomials

Last updated at May 29, 2023 by

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

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Ex 2.4, 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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