<s>[INST] Eliminating $x$, $y$, $z$ from $\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$ and $ax+by+cz=0$ Here is a Math Tripos problem I cannot solve.

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

I dont know if there is a general (practical) method for such problem. I think you just manipulate the equations to eliminate $x,y,z$ but have been unable to find the path in this question.
The answer is $a^3+b^3+c^3=a^2(b+c)+b^2(a+c)+c^2(a+b)$ if that yields any insight.
 [/INST] 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
$$a^2(b+c)+b^2(c+a)+c^2(a+b) = a^3 + b^3+c^3$$
</s>