`|(2xy,x^2,y^2), (x^2,y^2,2xy), (y^2, 2xy, x^2)|=` (A) `(x^3+y^3)^2` (B) `(x^2+y^2)^3` (C) `-(x^2+y^2)^3` (D) `-(x^3+y^3)^2`

HCF of x^2 - y^2 and x^3 - y^3 is (A) x-y (B) x^3 -y^3 (C ) (x^2 - y^2) (D) (x+y ) (x^2 xy + y^2)

x^(3) + x^(2)y+ xy^(2) + y^(3) = 81

solve: -2 (x^(2) - y^(2) + xy) - 3 (x^(2) + y^(2) - xy)

Add (2x^2+5xy+2y^2), (3x^2-2xy+5y^2) and (-2x^2+8xy+3y^2)

Add : x^(3) - x^(2)y + 5xy^(2) + y^(3) , -x^(3) - 9xy^(2) + y^(3), 3x^(2)y + 9xy^(2)

Verify the following : (i) x^3+y^3=(x-y)(x^2-xy+y^2) (ii) x^3-y^3=(x-y) (x^2+xy+y^2)

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^2 + xy + 3y^2 = 1 then (x + 6y)^3 (d^2y)/(dx^2) is