已知a⼀(x2-yz)=b⼀(y2-zx)=c⼀(x2-xy) 求证ax+by+cz=(x+y+z)(a+b+c) 我要过程，那种看不懂的别抄过来

a/(x2-yz)=b/(y2-zx)=c/(x2-xy) ＝ k (1)

利用合分比
(a+b+c) / (x^2+y^2+c^2-xy-yz-xz) = k (2)
(1)左边分别在分子，分母乘以a,b,c得
ax/x^3-xyz=by/(y^3-xyz)=cz/(x^3-xyz) = k
利用合分比
=(ax+by+cz) / (x^3+y^3+z^3-3xyz) = k
由于x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+c^2-xy-yz-xz) = (x+y+z) * (a+b+c)/k 将（2）代入
所以ax+by+cz = k(x^3+y^3+z^3-3xyz) = k (x+y+z)(a+b+c) / k = (x+y+z)(a+b+c)

令a/(x2-yz)=b/(y2-zx)=c/(x2-xy)=k，
得a=(x2-yz)k.......... (1)
b=(y2-zx)k;............ (2)
c=(x2-xy)k............. (3)
将ax+by+cz=(x+y+z)(a+b+c)左边乘开 化简为a(y+z)+b(x+z)+c(x+y)=0
(1)*（y+z）+(2)*(x+z)+(3)*(x+y)=0
得证