Do You Prefer Quadratics To Cubics?

Relevant wiki: Vieta's Formula - Higher Degrees

Rearranging the equations, we have:

$a^3-a-1=0$

$b^3-b-1=0$

$c^3-c-1=0$

Which clearly means that $a,b,$ and $c$ are three distinct roots of the polynomial $x^3-x-1=0$ . (distinct because the polynomial cannot be factorised in a way such that it has atleast two common factors)

The polynomial can also be expressed as $x^3 + (0)x^2 + (-1)x + (-1)=0$

Then clearly from Vieta's Formulas,

$a+b+c=0$

$ab+bc+ca=-1$

$abc=1$

Now,

$a^3+b^3+c^3$

$=(a+1)+(b+1)+(c+1)=(a+b+c)+3=0+3=\boxed{3}$

By adding the equations together you get a^3 + b^3 + c^3 = a + b + c +3

We see (as others have stated in better formats), that a, b and c are the distinct roots of x^3 - 0x^x - x -1 = 0.

By Vieta's Formula, the sum of the roots is (0/1) = 0. Therefore a + b + c + 3 = 0 + 3 = 3.