a, b, c are distinct non-zero reals such that $$\frac{1+a^3}{a} = \frac{1+b^3}{b} = \frac{1+c^3}{c}$$. Find all possible values of $$a^3+b^3+c^3$$.

Setting all three expressions equal to a common value k shows that a, b and c are all roots of the cubic $$x^3-kx+1=0$$. Since this cubic has no $$x^2$$ term, the sum of its roots is $$a+b+c=0$$, and its constant term gives $$abc=-1$$. Using the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$, the right side vanishes since $$a+b+c=0$$, so $$a^3+b^3+c^3=3abc=3(-1)=-3$$.