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Algebra Difficulty 4.8 AIME Find the answer

Example 9 If x,y,zx, y, z are real numbers, and
(y−z)2+(z−x)2+(x−y)2=(y+z−2x)2+(z+x−2y)2+(x+y−2z)2, \begin{aligned} (y-z)^{2} & +(z-x)^{2}+(x-y)^{2} \\ = & (y+z-2 x)^{2}+(z+x-2 y)^{2} \\ & +(x+y-2 z)^{2}, \end{aligned}

find the value of M=(yz+1)(zx+1)(xy+1)(x2+1)(y2+1)(z2+1)M=\frac{(y z+1)(z x+1)(x y+1)}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution: The condition can be simplified to
x2+y2+z2−xy−yz−zx=0. x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .

Then (x−y)2+(y−z)2+(z−x)2=0(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0,
which implies x=y=zx=y=z.
Therefore, M=1M=1.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.