$(32)^3 + (79)^3 - (111)^3 + 3 \times 32 \times 79 \times 111$ is equal to

### Concept & Formula This expression evaluates to zero based on a clever disguised application of the conditional identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 - 3abc = 0$. ### Step-by-Step Solution * Look closely at the structure of the expression: $(32)^3 + (79)^3 + (-111)^3 - 3(32)(79)(-111)$. * Notice that the negative signs have been distributed. The expression exactly matches the form $a^3 + b^3 + c^3 - 3abc$. * Let $a = 32$, $b = 79$, and $c = -111$. * Note that in the given expression, subtracting $(111)^3$ is mathematically identical to adding $(-111)^3$. * Similarly, adding $+ 3 \times 32 \times 79 \times 111$ is identical to subtracting $-3 \times (32) \times (79) \times (-111)$. * Check the sum of the variables: $$a + b + c = 32 + 79 + (-111) = 111 - 111 = 0$$ * Because $a + b + c = 0$, the entire expression $a^3 + b^3 + c^3 - 3abc$ must equal $0$. ### Exam Strategy & Shortcut Any time an aptitude question asks you to add and subtract massive cubed numbers alongside a product of those same numbers, it is **always** testing the $a+b+c=0$ rule. Don't calculate anything; just verify that the base numbers sum to zero. ### Common Pitfall Failing to recognize that $c$ is negative ($-111$) and incorrectly assuming the formula doesn't apply because the signs look slightly different than the standard textbook formula. ### Final Answer Therefore, the correct answer is **0**.