Can you help me solve this algebra problem? Hi I need to solve this problem and I don’t know how. I’d appreciate your help. 

If $x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}$ and $x\neq y\neq z$, then $$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$

I think I need to 
$x^3 - ayz = x^2k$
$y^3 - azx = y^2k,$
$z^3 - axy = z^2k$ 
then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.
 A: Denote $k=xyz$ and $b$ the common value of $x-\frac{ayz}{x^2}=y-\frac{azx}{y^2}=z-\frac{axy}{z^2}$. We can see that the equation
$$
t-\frac{ak}{t^3}=b\tag{*}
$$ is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
$$
t^4-b t^3-ak=0.
$$ By Vieta's formula, the other root $w$ satisfies
$$
x+y+z+w=b,\quad xyzw=-ak.
$$ Since $k=xyz\ne 0$ it follows that
$$
w=-a=b-x-y-z,
$$ hence we get
$$
x-\frac{ayz}{x^2}= y-\frac{azx}{y^2}=z-\frac{axy}{z^2}=b=x+y+z-a.
$$
A: $$x-\frac{ayz}{x^2}=y-\frac{axz}{y^2}$$ gives
$$x-y+az\left(\frac{x}{y^2}-\frac{y}{x^2}\right)=0$$ or
$$1+\frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
Similarly, $$\frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
$$\frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
$$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
$$\sum_{cyc}(x^2y^2+x^2yz)=0$$ or
$$\sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
$$x=y=z=0,$$ which is impossible. 
Id est, the given is wrong, which says that
$$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}\Rightarrow x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$ is true.
