Grade 12Algebra
Evaluate using identities..(99)3Please tell me easy way to do it
Evaluate using identities..
(99)3
Please tell me easy way to do it
1 Answer
Arun
Dear Priya
It 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
(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
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algebra
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