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Let $$k=\frac{a^3+6}{a}=\frac{b^3+6}{b}=\frac{c^3+6}{c}.$$Then $a,$ $b,$ and $c$ are all roots of $$k=\frac{x^3+6}{x},$$or $x^3-kx+6=0.$ By Vieta's formulas, $a+b+c=0.$

Also, $$\begin{array}{rl}{a}^3-ka+6& =0,\\ b^3-kb+6& =0,\\ c^3-kc+6& =0.\end{array}$$Adding these, we get $a^3+b^3+c^3-k(a+b+c)+18=0,$ so $a^3+b^3+c^3=k(a+b+c)-18=\boxed{-18}.$