Ex 2.5, 7 - Evaluate using suitable identities (i) (99)3 - Identity VI & VII

 

 

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

Transcript

Ex 2.5, 7 Evaluate the following using suitable identities: (i) (99)3 We write 99 = 100 – 1 (99)3 = (100 − 1)3 Using (a – b)3 = a3 – b3 – 3ab(a – b) Where a = 100 & b = 1 = (100)3 − (1)3 − 3(100) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299

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