x3-xyz=2y3-xyz=6z3-xyz=20,求方程组的实根

Question

x3-xyz=2 y3-xyz=6 z3-xyz=20,求方程组的实根

Answer 1

令t=xyz,则
x^3=t+2
y^3=t+6
z^3=t+20
三式相乘：
(xyz)^3=(t+2)(t+6)(t+20),即
t^3=t^3+28t^2+172t+240
即7t^2+43t+60=0
(7t+15)(t+4)=0
得：t=-15/7 or -4
因此有两组实根：
当t=-15/7,x=(t+2)^(1/3)=-1/7^(1/3),y=(t+6)^(1/3)=(28/7)^(1/3),z=(t+20)^(1/3)=(125/7)^(1/3)
当t=-4,x=(t+2)^(1/3)=-2^(1/3),y=2^(1/3),z=16^(1/3)