Without actually calculating the cubes, find the value of each of the following:
$(28)^{3} + (- 15)^{3} + (- 13)^{3}$

17530

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14850

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12022

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16380

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Solution

The correct options are
A 17530
C 16380
We know that,
$I f (a + b + c) = 0, t h e n a^{3} + b^{3} + c^{3} = 3 a b c$
Here, $28 + (- 15) + (- 13) = 0$
$∴ (28)^{3} + (- 15)^{3} + (- 13)^{3} = 3 (28) (- 15) (- 13) = 16380$

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