<s>[INST] Use of property of ratio and factorization method Recently I have faced a problem related to algebraic ratio.I have tried much but I can not find any clue.Can anyone give me some hints.
Here is the problem.
I am given $$\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\neq0$$
I have to prove: $(a+b+c)(x+y+z)=ax+by+cz$
Please help me to solve the problem with clear explanation
 [/INST] Define $k$ by $$\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}=k\neq0,$$
so $1/k\ne0$.
Now, finding the product, we have
$$\begin{align}(a+b+c)(x+y+z)=&\;ax+by+cz\\&+a(y+z)\\ &+b(x+z)\\&+c(x+y);\end{align}$$
thus we need to show that $a(y+z)+b(x+z)+c(x+y)=0$.
Since
$$\frac{x^2-yz}{k}\cdot(y+z)=a(y+z)\,$$
$$\frac{y^2-zx}{k}\cdot(x+z)=b(x+z)\,$$
$$\frac{z^2-xy}{k}\cdot(x+y)=c(x+y),$$
we obtain
$$\begin{align}a(y+z)+b(x+z)+c(&x+y)\\&=\frac1k((x^2-yz)(y+z)\\&\qquad+(y^2+zx)(x+z)\\&\qquad+(z^2+xy)(x+y))\\&=0.\end{align}$$
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