Ex 2.4, 7 (i) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 13, 2024 by Teachoo
Ex 2.4
Ex 2.4, 1 (i)
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Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
- Ex 2.3
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Ex 2.4
- Examples
- Remainder Theorem
Transcript
Ex 2.4, 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
