Question

x3-y3-x+y plz help me ​

Answer 1

Answer:

(X-Y)³=X³-Y³-3XY(X-Y)

so

(X-Y)³+3XY(X-Y)-(X-Y)=0

(X-Y)((X-Y)²+3XY-1)=0

now you can solve it by quadratic equation in 2 degree

for 1st

X=Y

and by putting X=Y

X=sqr. root 1/3