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)

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)
