Problem:
Let (a,b,c) be a real solution of the system of equations
x3−xyz=2y3−xyz=6z3−xyz=20
The greatest possible value of a3+b3+c3 can be written in the form nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Add xyz to each side of each equation to obtain x3=2+xyz,y3=6+xyz, and z3=20+xyz. Letting P=xyz, it follows that P3=(2+P)(6+P)(20+P)=P3+28P2+172P+240. Simplifying yields the equation 7P2+43P+60=0, and thus P=−715 or P=−4. By adding the original three equations, it follows that x3+y3+z3=28+3P, and this can be maximized by taking the greater of the two values of P, which is −715. This value of P corresponds to the solution (−371,373,375) of the system. Thus the largest possible value of x3+y3+z3 is 28−745=7151, and m+n=151+7=158.
Note: The solution corresponding to P=−4 is (−32,32,232).