<s>[INST] $\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}, a \neq b \neq c$; What is $a^3+b^3+c^3$? 
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}\\a \neq b \neq c\\a^3+b^3+c^3=$

Since $\frac{1-a^3}{a}=\frac{1-b^3}{b}$, $a^3=b^2a-\frac{a}{b}-1$.
$\therefore a^3+b^3+c^3=ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1$
Since $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$,
$ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
(stuck here)
 [/INST] $a, b, c$ are roots of $$\frac{1-x^3}{x}=m$$
$$x^3+mx-1=0$$
Hence, we know that $a+b+c=0$ and $abc=1$ from Vieta's formula. Now you can use the identity that you stated to solve for $a^3+b^3+c^3$.
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