Problem:
Let x and y be real numbers satisfying x4y5+y4x5=810 and x3y6+y3x6=945. Evaluate 2x3+(xy)3+2y3.
Solution:
Note that neither x nor y can equal zero, as otherwise, the left-hand sides of the two given equations would both equal 0. Therefore let y=kx for some nonzero value of k. The given equations then become
x9k5+x9k4=810 and x9k6+x9k3=945
Note that the left-hand sides of the above equations have a common factor of x9k3(k+1). Furthermore, k cannot equal −1, as otherwise, the left-hand sides of the above two equations would both equal 0. Thus
810945=67=x9k5+x9k4x9k6+x9k3=x9k4(k+1)x9k3(k+1)(k2−k+1)
which simplifies to 6k2−13k+6=0. The solutions of this quadratic equation are k=32 and 23. Because k is positive, it follows that x and y must also be positive. When k=32,x9⋅(24332+8116)=810, so x9=80810⋅243=2339. Then x3=227 and y3=(32)3⋅227=4. Similarly, if k=23, then x3=4 and y3=227. In either case, 2x3+(xy)3+2y3=2⋅227+227⋅4+2⋅4=89.
The problems on this page are the property of the MAA's American Mathematics Competitions