I want to know the best answer of the question from Polynomials chapter of class 9th ncert math. The question from exercise 2.5 of math. Give me the easy way for solving this question of 7(3) Evaluate the following using suitable identities: (iii) (998)3
Share
We can write 99 as 1000–2
Using identity,(x–y)3 = x3–y3–3xy(x–y)
(998)3 =(1000–2)3
=(1000)3–23–(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8- 6000000+12000
= 994011992