symbolic - Efficient code for solve this equation

We have $a*b*c=-1$, $\frac{a^2}{c}+\frac{b}{c^2}=1$, $a^2 b+a c^2+b^2 c=t$

What's the value of $a^5 c+a b^5+b c^5$?

I tried

Eliminate[{a b c == -1, a^2/c + b/c^2 == 1, a^2 b + b^2 c + c^2 a == t, 
  a b^5 + b c^5 + c a^5 == res}, {a, b, c}]

It's much slower than Maple's eliminate. How do I solve this efficiently?

Answer

If you use the third argument in Solve, i.e. a list of variables to be eliminated (take a look at the Eliminating Variables tutorial in Mathematica) then you'll get the result immediately :

Solve[{a b c == -1, a^2/c + b/c^2 == 1, 
       a^2 b + b^2 c + c^2 a == t,
       a b^5 + b c^5 + c a^5 == res},
      {res}, {a, b, c, t}]

{{res -> 3}}

Edit

It should be underlined that Solve appears to be smarter than Eliminate due to its improvement in Mathematica 8, look at its options, e.g. MaxExtraConditions, Method ( Method -> Reduce). However most of the update of Solve is hidden, but in general it shares its methods with Reduce. Defining

system = { a b c == -1,
           a^2/c + b/c^2 == 1, 
           a^2 b + b^2 c + c^2 a == t,
           a b^5 + b c^5 + c a^5 == res };

then it works too

Solve[ system, {res}, {a, b, c}]

{{res -> 3}}

while it doesn't in Mathematica 7 yielding

No more memory available.
Mathematica kernel has shut down.
Try quitting other applications and then retry.

and your original problem should be evaluated this way (you've lost t):

Eliminate[ system, {a, b, c, t}]

res == 3

and it works in Mathematica 7 as well.