**Question:**

If (x^2)/by+cz=y^2/cz+ax=z^2/ax+by=1,then value ofa/a+x +b/b+y +c/c+z is 1,-1,2,-2

So, x^2+y^2+z^2=2(ax+by+cz)

Or, (x-a)^2+(y-b)^2+(z-c)^2=a^2+b^2+c^2

On compairing, we get

x=2a, y=2b and z=2c

Now, a/(a+x)+b/(b+y)+c/(c+z)

Putting the value of x, y and z we get

1/3+1/3+1/3=1 ans.

Yes

Do you need explaination?

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