I want to know the best solution of the question from class 9th ncert math of Polynomials chapter of exercise 2.5 of math How i solve the question of this class in simple and easy way Evaluate the following using suitable identities: (i) (99)3
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We can write 99 as 100–1
Using identity, (x –y)3 = x3–y3–3xy(x–y)
(99)3 = (100–1)3
= (100)3–13–(3×100×1)(100–1)
= 1000000 –1–300(100 – 1)
= 1000000–1–30000+300
= 970299