Evaluate the following using suitable identities: (a) (102)³,(b) (28)³+(-15)³+(-13)³
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b) Identity => a³+b³+c³=0
then, a³+b³+c³=3abc
Your answer.......
28-15-13=0
= then, 3abc= 3×28×-13×-15= 16380........
Hope it helps......:)
then, a³+b³+c³=3abc
Your answer.......
28-15-13=0
= then, 3abc= 3×28×-13×-15= 16380........
Hope it helps......:)
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