Evaluate \(24^3 + (-7)^3 + (-17)^3\)
8568
Take the three cube bases as \(a = 24\), \(b = -7\), \(c = -17\).
Check their sum: \(a + b + c = 24 - 7 - 17 = 0\).
When \(a + b + c = 0\), the identity \(a^3 + b^3 + c^3 = 3abc\) applies.
So \(24^3 + (-7)^3 + (-17)^3 = 3 \times 24 \times (-7) \times (-17)\).
Evaluate: \(3 \times 24 \times 119 = 72 \times 119 = 8568\) (the two negatives multiply to \((-7)\times(-17) = 119\)).
Hence, the value of the expression is 8568.
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