Showing $(a+b+c)(x+y+z)=ax+by+cz$ given other facts $$x^2-yz/a=y^2-zx/b=z^2-xy/c$$
None of these fractions are equal to 0.We need to show that,
$(a+b+c)(x+y+z)=ax+by+cz$ 
This question comes from a chapter that wholly deals with factoring homogeneous cyclic polynomials.I multiplied the three sides of the  first equality by $abc$ but that yields an unfactorizable polynomial.I haven't had much luck in manipulating the first equality.So I tried to understand what I was trying to prove by expanding the second equality.We are trying to prove that,
$$a(y+z)+b(x+z)+c(x+y)=0$$
But the LHS is still unfactorizable.I tried to manipulate the first equality more,but they yielded nothing.However,I did find the following equality:
$$bc(x^2-yz)-ac(y^2-zx)+ab(z^2-xy)=a(y+z)+b(x+z)+c(x+y) $$ 
I would appreciate a very small hint.

Using Addendo formula $\displaystyle \frac Aa=\frac Bb=\cdots=\frac{A+B+\cdots}{a+b+\cdots},$
$$\frac{x^2-yz}a=\frac{y^2-zx}b=\frac{z^2-xy}c=\frac{x^2-yz+y^2-zx+z^2-xy}{a+b+c}\ \ \ \ (1)$$
Multiplying the numerator  & the denominator of the first term by $x,$ and those of second by $y$  and those of third by $z$
$$\implies\frac{x^3-xyz}{ax}=\frac{y^3-xyz}{by}=\frac{z^3-xyz}{cz}=\frac{x^3+y^3+z^3-3xyz}{ax+by+cz}\ \ \ \ (2)\text{ (again applying Addendo)}$$
Equate $(1),(2)$ using Factorize the polynomial $x^3+y^3+z^3-3xyz$
