Let \(a, b,\) and \(c\) be distinct real numbers such that \(\frac{a^3 + 6}{a} = \frac{b^3 + 6}{b} = \frac{c^3 + 6}{c}.\) Find \(a^3 + b^3 + c^3.\)
By inspection:
\(\frac{1^3+6}{1}=7\)
\(\frac{2^3+6}{2}=7\)
Unfortunately \(\frac{3^3+6}{3}\ne7\)
However \(\frac{(-3)^3+6}{-3}=7\)
I'll leave you to finish.
tysm!!! the answer is -18 for anyone wondering :)
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