Thank you very much!
Yes, I had tried it before. You get:
x^3 + xy^2 + x^2*y + y^3 + 1 = 0
What about now? I tried to obtain two lower-grade factors with Ruffini's rule, but I can't find a root for the expression.
Don't forget that you divided by x- y to get that. So x= y is a separate solution. If x= y, what must z be?
Also, if you subtract the third equation from the first, x- z= 0 or [itex]x^3+ zx^2+ xz^2+ z^4+ 1= 0[/itex]. Use that with your equation in x and y.
When you divide by y - x you do not get 0 on the RHS, you get 1. I think this is causing some of your confusion.
x -y is a solution, but not the only solution.
Let's try factoring instead of division:
First let's put things in an easier format:
Note the change I'm making on the RHS:
Now factor:
So we know that either
or
The first equation gives y = x. The other equation? I'm not telling. Give it a try.
-Dan