Solve the system of equations over real numbers: ⎩⎨⎧x3=2y3+z−2,y3=2z3+x−2,z3=2x3+y−2.
Solution
From the first two equations of our system we obtain: x3−y3=2y3−2z3+y−z=(y−z)(2y2+2yz+2z2+1), analogously, we can easily get equalities: y3−z3=(z−x)(2z2+2zx+2x2+1), z3−x3=(x−y)(2x2+2xy+2y2+1). It is easy to see that (2y2+2yz+2z2+1)=y2+z2+(y+z)2+1>0. If we suppose that y>z, then from the first equality we have that x>y, then applying third equality we conclude: z>x>y>z - contradiction. The case when y<z is completely analogous. Hence we obtain x=y=z. and solving the equation we get the answer (1,1,1).
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