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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 $a^3 + b^3 + c^3.$

Alan · Answer 1 · 2022-09-23T08:15:59

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.