2-y^(2)-y^(3)+2y^(8)

The following are the steps involved in factorizing 64 x^(6) -y^(6) . Arrange them in sequential order (A) {(2x)^(3) + y^(3)} {(2x)^(3) - y^(3)} (B) (8x^(3))^(2) - (y^(3))^(2) (C) (8x^(3) + y^(3)) (8x^(3) -y^(3)) (D) (2x + y) (4x^(2) -2xy + y^(2)) (2x - y) (4x^(2) + 2xy + y^(2))

If x-2y+2z=0, then x^(3)-8y^(3)+8z^(3) is equal to

If (x^(2) + y^(2))/(x^(2) - y^(2)) = 2(1)/(8) , find : (i) (x)/(y) (ii) (x^(3) + y^(3))/(x^(3) - y^(3))

Factorise: 12(y^(2)+7y)^(2)-8(y^(2)+7y)(2y-1)-15(2y-1)^(2)

The locus of a point "P" ,if the join of the points (2,3) and (-1,5) subtends right angle at "P" is x^(2)+y^(2)-x-8y+13=0 x^(2)-y^(2)-x+8y+3=0 x^(2)+y^(2)-4x-4y=0,(x,y)!=(0,4)&(4,0) x^(2)+y^(2)-x-8y+13=0,(x,y)!=(2,3)&(-1,5)

Show that the following set of curves intersect orthogonally: y=x^(3) and 6y=7-x^(2)x^(3)-3xy^(2)=-2 and 3x^(2)y-y=2x^(2)+4y^(2)=8 and x^(2)-2y^(2)=4