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

Find the products: (3x+2y)(9x^2-6x y+4y^2)

Given that x^2 + y^2 =4 and x^2 + y^2 -4x -4y =-4 , then x + y =

The equation of the transvers axis of the hyperbola (x-3)^2+(y+1)^2=(4x+3y)^2 is x+3y=0 (b) 4x+3y=9 3x-4y=13 (d) 4x+3y=0

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

The solution of the differential equation (dy)/(dx)+(x(x^(2)+3y^(2)))/(y(y^(2)+3x^(2)))=0 is (a) x^(4)+y^(4)+x^(2)y^(2)=c (b) x^(4)+y^(4)+3x^(2)y^(2)=c (c) x^(4)+y^(4)+6x^(2)y^(2)=c (d) x^(4)+y^(4)+9x^(2)y^(2)=c

Subtract: x^2-3x y+7y^2-2 from 6x y-4x^2-y^2+5

Consider the following equation in x and y: (x-2y-1)^2 + (4x+3y-4)^2 + (x-2y-1)(4x+3y-4) =0 How many solutions to (x, y) with x, y real, does the equation have

From the sum of x^2+3y^2-6x y ,2x^2-y^2+8x y , y^2+8 and x^2-3x y\ subtract -3x^2+4y^2-x y+x-y+3.

The equation of the circle which cuts the three circles x^2+y^2-4x-6y+4=0, x^2+y^2-2x-8y+4=0, x^2+y^2-6x-6y+4=0 orthogonally is

Find the product: (2x-y+3z)(4x^2+y^2+9z^2+2x y+3y z-6x z)