Finding $a^5 + b^5 + c^5$ 
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)

Given $a+b+c=3$ and $a^2+b^2+c^2 =5$ and $a^3+b^3+c^3=7$
Using $$ab+bc+ca = \frac{1}{2}\left[(a+b+c)^2-(a^2+b^2+c^2)\right] = 2$$
and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]=9$$
So $$7-3abc=9\Rightarrow abc=-\frac{2}{3}$$
Now Let $(t-a)\;,(t-b)\;,(t-c)$ be the root of cubic equation in terms of $t\;,$ Then
$$(t-a)(t-b)(t-c) =0\Rightarrow t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$
So $$t^3-3t^2+2t+\frac{2}{3}=0\Rightarrow t^4-3t^3+2t^2+\frac{2}{3}t=0......(1)$$
So $$\sum a^4-3\sum a^3+2\sum a^2+\frac{2}{3}\sum a=0$$, Where $\sum a^{n} = a^n+b^n+c^n\;,$ for $n=1,2,3,4,5$
So $$\sum a^4-3(7)+2\cdot 5+\frac{2}{3}\cdot 3=0\Rightarrow \sum a^4=9$$ 
Now $$t^5-3t^4+2t^3+\frac{2}{3}t^2=0$$ from equation $(1)$
So $$\sum a^5-3\sum a^4+2\sum a^3+\frac{2}{3}\sum a^2=0$$
So $$\sum a^5-3(9)+2(7)+\frac{2}{3}\cdot 5=0\Rightarrow \sum a^5 = \frac{29}{3}$$
