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 following steps are involved in expanding (x+ 3y)^(2) . Arrange them in sequential order from the first to the last . (A) (x + 3y)^(2) = x^(2) + 6xy + 9y^(2) (B) (x + 3y)^(2) = (x)^(2) + 2(x)(3y) + (3y)^(2) (C) Using the identify (a + b)^(2) = a^(2) + 2ab + b^(2) , where a = x and b = 3y.

Add: 8x^(2) - 5xy - 3y^(2), 2xy - 6y^(2) + 3x^(2) and y^(2) + xy - 6x^(2)

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

Find HCF of x^(3) + 3x^(2) y + 2xy^(2) and x^(4) + 6x^(3)y + 8x^(2) y^(2)

Add: 4x^(2) - 7xy + 4y^(2) - 3, 5 + 6y^(2) - 8xy + x^(2) and 6 - 2xy + 2x^(2) - 5y^(2)

(b) Factorize 4x^(3) - 10y^(3) - 8x^(2) y + 5xy^(2)

Factorise 25x^(2) - 30xy + 9y^(2) . The following steps are involved in solving the above problem . Arrange them in sequential order . (A) (5x - 3y)^(2) " " [ because a^(2) - 2b + b^(2) = (a-b)^(2)] (B) (5x)^(2) - 30xy + (3y)^(2) = (5x)^(2) - 2(5x)(3y) + (3y)^(2) (C) (5x - 3y) (5x - 3y)