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Algebra Difficulty 6.0 AIME, harder Prove it Greece

The real numbers x,y,zx, y, z, with x≠zx \neq z, are mutually different and nonzero and they satisfy the following equations:
(x+y)2+(2−xy)=9,(y+z)2−(3+yz)=4. (x+y)^2 + (2-xy) = 9, \\ (y+z)^2 - (3+yz) = 4.
Determine the value of the expression
A=(xy+y2x2+z3x2y)(yz+z2y2+x3y2z)(zx+x2z2+y3z2x). A = \left( \frac{x}{y} + \frac{y^2}{x^2} + \frac{z^3}{x^2y} \right) \left( \frac{y}{z} + \frac{z^2}{y^2} + \frac{x^3}{y^2z} \right) \left( \frac{z}{x} + \frac{x^2}{z^2} + \frac{y^3}{z^2x} \right).

Solution

The given equalities can be written:
x2+y2+xy=7,(1) x^2 + y^2 + xy = 7, \qquad (1)
y2+z2+yz=7,(2) y^2 + z^2 + yz = 7, \qquad (2)
By subtraction we get:
x2−z2+xy−yz=0⇔(x−z)(x+z)+y(x−z)=0⇔(x−z)(x+z+y)=0. x^2 - z^2 + xy - yz = 0 \Leftrightarrow (x-z)(x+z) + y(x-z) = 0 \Leftrightarrow (x-z)(x+z+y) = 0.
Since x−z≠0x - z \neq 0, we get:
x+y+z=0.(3) x + y + z = 0. \qquad (3)
Then with simple manipulations for the factors of AA we find
A=(x3+y3+z3xyz)3(4) A = \left( \frac{x^3 + y^3 + z^3}{xyz} \right)^3 \qquad (4)
From (3), using Euler's identity, we find
x3+y3+z3=3xyz(5) x^3 + y^3 + z^3 = 3xyz \qquad (5)
Hence from (4) and (5) we get A=27A = 27

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