> 
> Suppose we have numbers $a,b,c$ which satisfy the equations
>  $$a+b+c=3,$$
>  $$a^2+b^2+c^2=5,$$
>  $$a^3+b^3+c^3=7.$$
> 
> 
> 

How can I find $a^5 + b^5 + c^5$?

I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:

$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac{2}{3} \rangle$$

I solved the last equation for $c$ and got 3 complex values. When I plug into the 2nd equation $(b^2+bc+c^2-3b-3c+2)$ I get a lot of roots for $b$, and it is laborious to plug in all these values.

Is there a shortcut or trick to doing this? The hint in the book says to use remainders. I computed the remainder of $f = a^5 + b^5 + c^5$ reduced by $G$:
$$\overline{f}^G = \frac{29}{3}$$

How can this remainder be of use to me?

Thanks. (Note: I am using Macaualay2)