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

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Solution

${(28)}^{3}$ + ${(- 15)}^{3}$ + ${(- 13)}^{3}$

$∵$ (28) + (-15) + (-13) = 0

Using identity, if x + y + z = 0, then $x^{3}$ + $y^{3}$ + $z^{3}$ = 3xyz.

= 3 × (28) × (-15) × (-13) = 16380 ...(answer)

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