$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}, a \neq b \neq c$; What is $a^3+b^3+c^3$?

Question

$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}\\a \neq b \neq c\\a^3+b^3+c^3=$

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)

Answer 1

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

Answer 2

It looks like you don't need to do any calculations.

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We can construct a new cubic equation with roots $a^3,b^3,c^3.$

$$a^3=x_1,~ b^3=x_2,~ c^3=x_3$$

$$\begin{align}r=\frac{1-a^3}{a}=\frac{1-b^3}{ b}=\frac{1-c^3}{c}\end{align}$$

$$\begin{align}&\implies1-a^3=ar \\ &\implies(1-a^3)^3=a^3r^3 \\ &\implies(1-x)^3=xr^3 \\ &\implies1-3x+3x^2-x^3-xr^3=0 \\ &\implies x^3-3x^2+x(r^3+3)-1=0 \\ &\implies x_1+x_2+x_3=3 \\ &\implies a^3+b^3+c^3=3.\end{align}$$