Example 18 (ii) - Chapter 2 Class 9 Polynomials
Last updated at Dec. 16, 2024 by Teachoo
Examples
Example 1 (i)
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Chapter 2 Class 9 Polynomials
Serial order wise
- Ex 2.1
- Ex 2.2
- Ex 2.3
- Ex 2.4
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Examples
- Remainder Theorem
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
