" (iv) "[x-(2)/(3)y]^(3)" .anfsic "

On simplification, (x^3-y^3)/(x[(x+y)^2-3xy]) div (y[(x-y)^2+3xy])/(x^3+y^3)xx ((x+y)^2-(x-y)^2)/(x^2-y^2) is equal to : सरलीकरण के बाद, (x^3-y^3)/(x[(x+y)^2-3xy]) div (y[(x-y)^2+3xy])/(x^3+y^3)xx ((x+y)^2-(x-y)^2)/(x^2-y^2) किसके बराबर होगा?

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))

The value of the expression ((x ^(2) - y ^(2)) ^(3) + ( y ^(2) - z ^(2)) ^(3) + (z ^(2) - x ^(2)) ^(3))/((x - y) ^(3) + ( y - z) ^(3) + (z - x ) ^(3)) is

If y = 2x + 3x ^ (2) + 4x ^ (3) + ......., then (y) / (2) - (1.3) / (2!) ((Y) / (2) ) ^ (2) + (1.3.5) / (3!) ((Y) / (3)) ^ (3) -...... oo =

The solution of (dy)/(dx) = (3x^(2)y - y^(3))/(3y^(2) x -x^(3)) is

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

Find ((x^(2)-y^(2))^(3) + (y^(2) -z^(2))^(3)+ (z^(2) -x^(2))^(3))/((x-y)^(3) + (y-z)^(3) + (z-x)^(3))