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?
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.
Can compute a Groebner basis with an ordering that eliminates {a,b,c}.
eqns = {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}; GroebnerBasis[ Numerator[Together[Apply[Subtract, eqns, {1}]]], {res, t}, {a, b, c}] (* Out[150]= {-3 + res} *) The result is now immediate.
I found two methods:
Reduce[{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}, {t}] // First (*res == 3*) res /. Solve[{a b c == -1, a^2 + b/c == c, a^2 b + b^2 c + c^2 a == t, a b^5 + b c^5 + c a^5 == res}, {a, b, c, res}] // Union (*{3}*) res /. {ToRules @ Reduce[{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}, {t}]}. We can observe that we do not need this equation : a^2 b + b^2 c + c^2 a == t, i.e. we can get the solution this way res /. {ToRules @ Reduce[{a b c == -1, a^2/c + b/c^2 == 1, a b^5 + b c^5 + c a^5 == res}, {res}]}. $\endgroup$ Normal @ Solve[{a b c == -1, a^2/c + b/c^2 == 1, a b^5 + b c^5 + c a^5 == res}, {res}, MaxExtraConditions -> All] $\endgroup$