Solution :
Given x³-4x²y-xy²+4y³=0
(x-4y)(x-y)(x+y)=0
the three line on (x,y) plain are
y=-x
y=x
y=x/4
a) Find the real values of x for which
(x^3)-4(x^2)-x+4>=0, which I got to be x>=4 and -1<=x<=1. That's right, right?
It's the next part that I'm really stuck with
b) Find the three lines in the (x,y) plane on which
(x^3)-4(x^2)y-x(y^2)+4(y^3)=0.
I'll be really grateful for your help.