Monday Math 72

What are the conditions on real numbers x, y, z such that ?

Well, inspection shows us that if (x,y,z) is a solution, so are the other 5 permutations of that solution. We also note that if y=-x, we see that , and , so the equality holds for y=-x, and thus, via the permutations, for z=-x and z=-y. However, are there any other solutions? Some algebra allows us to check. We will use the difference of cubes formula and the sum of cubes formula . So:
Moving the terms to the same side,

So our equality is true if x+y=0, x+z=0, or y+z=0; precisely the solution we found by inspection.


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