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Evaluate by using suitable identitiesa) [99]³b)101×99 |
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Answer» t is known that, (a+b)3=a3+b3+3ab(a+b)and(a−b)3=a3−b3−3ab(a−b) (i) (99)3 = (100 − 1)3 = (100)3 − (1)3 − 3(100) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299 (ii) (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100) (2) (100 + 2) = 1000000 + 8 + 600 (102) = 1000000 + 8 + 61200 = 1061208 (iii) (998)3= (1000 − 2)3 = (1000)3 − (2)3 − 3(1000) (2) (1000 − 2) = 1000000000 − 8 − 6000(998) = 1000000000 − 8 − 5988000 = 1000000000 − 5988008 = 994011992Step-by-step explanation: |
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