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)
 A: You have to use Newton Identities. See https://en.wikipedia.org/wiki/Newton%27s_identities
In general if you have $n$ variables $x_1\ldots.x_n$, define the polynomials
$$p_k(x_1,\ldots,x_n)=\sum_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$$
and
 \begin{align}
  e_0(x_1, \ldots, x_n) &= 1,\\
  e_1(x_1, \ldots, x_n) &= x_1 + x_2 + \cdots + x_n,\\
  e_2(x_1, \ldots, x_n) &= \textstyle\sum_{1\leq i<j\leq n}x_ix_j,\\
  e_n(x_1, \ldots, x_n) &= x_1 x_2 \cdots x_n,\\
  e_k(x_1, \ldots, x_n) &= 0, \quad\text{for}\ k>n.\\
 \end{align}
Then
\begin{align}
  p_1 &= e_1,\\
  p_2 &= e_1p_1-2e_2,\\
  p_3 &= e_1p_2 - e_2p_1 + 3e_3 ,\\
  p_4 &= e_1p_3 - e_2p_2 + e_3p_1 - 4e_4, \\
      & {}\ \ \vdots\\
\\
  e_0 &= 1,\\
  e_1 &= p_1,\\
  e_2 &= \frac{1}{2}(e_1 p_1 - p_2),\\
  e_3 &= \frac{1}{3}(e_2 p_1 - e_1 p_2 + p_3),\\
  e_4 &= \frac{1}{4}(e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4),\\
      & {} \  \  \vdots
\end{align}
In your case you have only 3 variables. Using the formulas above compute
\begin{align}
e_1 &=p_1=3, \\
e_2 &=2,\\
e_3 &=-\frac{2}{3}\\
e_4 &=0\\
e_5 &=0\\
\end{align}
Then compute $p_4=9$ and $$p_5=\frac{29}{3}.$$
A: Using just Macaulay2, you can do the following
Macaulay2, version 1.6.0.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
               ReesAlgebra, TangentCone

i1 : R=QQ[a,b,c]

o1 = R

o1 : PolynomialRing

i2 : i1=ideal(a+b+c-3,a^2+b^2+c^2-5,a^3+b^3+c^3-7)

                            2    2    2       3    3    3
o2 = ideal (a + b + c - 3, a  + b  + c  - 5, a  + b  + c  - 7)

o2 : Ideal of R

i3 : S=R/i1

o3 = S

o3 : QuotientRing

i4 : phi=map(S,R)

o4 = map(S,R,{- b - c + 3, b, c})

o4 : RingMap S <--- R

i6 : use R

o6 = R

o6 : PolynomialRing

i7 : phi(a^5+b^5+c^5)

     29
o7 = --
      3

o7 : S

(I deleted i5 and o5 as I made a typo in the input there)
A: 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 $$\sum a^5-3(9)+2(7)+\frac{2}{3}\cdot 5=0\Rightarrow \sum a^5 = \frac{29}{3}$$
