Question

求解一个三元一次方程

Answer 1

如下

Answer 2

f(x) = x^3-3k^2x+2k^3
f(k)=0
x^3-3k^2.x+2k^3 =(x-k)(x^2+ax -2k^2)
coef. of x
-2k^2-ka = -3k^2
ka = k^2
a = k
//
x^3-3k^2.x+2k^2 =0
(x-k)(x^2+kx -2k^2)=0
(x-k)(x+2k)(x-k)=0
x=k or -2k