If $\ \frac{a^3+1}{a} = \frac{b^3+1}{b} = \frac{c^3+1}{c} $ find $\ a^3+b^3+c^3 $ if a,b and c are distinct

Question

I know that this question can be solved using Vieta's relations somehow but I don't know how. I tried solving it with basic algebra but to no avail. I tried to make an equation with roots a,b and c but I didn't get anywhere with that either.

I cross multiplied the first two equations and did some factoring to get $(a-b)[ab(a+b)-1] = 0.$ Hence, $a=b$ or $ab(a+b)=1.$

I also tried making an equation $x^3 -px^2 +qx -r=0$ with roots a,b and c. But my attempts reached dead ends. I'd love to see your solution.

Answer 1

You got that $$ab(a+b)=1.$$ By the same way:$$ac(a+c)=1$$ and $$bc(b+c)=1.$$ Thus, $$ab(a+b)=ac(a+c)$$ or $$ab+b^2-ac-c^2=0$$ or $$a(b-c)+(b-c)(b+c)=0$$ or $$(b-c)(a+b+c)=0$$ or $$a+b+c=0.$$ Note that b and c are distinct, so dividing by (b-c) is valid as $b-c≠0$ Id est, $$a^3+b^3+c^3=a^3+b^3-(a+b)^3=-3ab(a+b)=-3.$$

Answer 2

Hint: $a, b, c$ are the three distinct roots to $ x^3 - k x + 1 = 0$.

So $ a+b+c = 0, ab+bc+ca = -k, abc = -1$ from Vietas.
There are a few ways to complete this from here.