Eliminating $x$, $y$, $z$ from $\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$ and $ax+by+cz=0$

Let's denote $t$ be equal to

$$t = \frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$

We have: \begin{align} t &=\frac{(y^2-yx-yz)+(z^2-zx-zy)-(x^2-xy-xz)}{b+c-a} \\ &=\frac{(y^2-2yz+z^2)-x^2}{b+c-a} \\ &=\frac{(y-z)^2-x^2}{b+c-a} \\ &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \\ \end{align} Hence, \begin{align} t^2 &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \times \frac{x^2-xy-xz}{a} \\ &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \times \frac{-x(y+z-x)}{a} \\ &=(x+y-z)(y+z-x)(z+x-y) \times \frac{x}{a(b+c-a)} \\ \end{align}

We have then $$\frac{a(b+c-a)}{x} = \frac{b(c+a-b)}{y} = \frac{c(a+b-c)}{z} = u $$ $$\left(\text{both equal to } u =\frac{(x+y-z)(y+z-x)(z+x-y)}{t^2} \right)$$

We can deduce that $$u = \frac{a^2(b+c-a)}{ax} = \frac{b^2(c+a-b)}{by} = \frac{c^2(a+b-c)}{cz} $$ Hence: $$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c) = u(ax+by+cz) = 0$$ (because $ax +by +cz = 0$)

Finally, we can conclude that $$a^2(b+c)+b^2(c+a)+c^2(a+b) = a^3 + b^3+c^3$$