(6)^(2)2x-3y=4;3y-x=4

(6) 2x-3y=4 ; 3y-x=4

2x+3y=5 4x-5y=6

2x+3y=8 4x-5y=-6

Show that (x^(2)+y^(2))^(4)=(x^(4)-6x^(2)y^(2)+y^(4))^(2)+(4x^(3)y-4xy^(3))^(2)

The pole of 2x+3y=0 with respect to x^(2)+y^(2)+6x-4y=0 is

4x-2y=3 6x-3y=5

Divide: 3y^(4)-3y^(3)-4y^(2)-4y^(2)-4y-byy^(2)-2y2y^(5)+10y^(4)+6y^(3)+y^(2)+5y+3by2y^(3)+1x^(4)-2x^(3)+2x^(2)+x+4byx^(2)+x+1

Simplify : (x+y)^(4) - 4y(x+y)^(3) + 6y^(2)(x+y)^(2)-4y^(3)(x+y)+y^(4)