If `(x+y)^3-\ (x-y)^3-6y\ (x^2-y^2)=\ k y^3,` then `k=` (a)1 (b) 2 (c) 4 (d) 8

If 2x+3y=5 and 3x+2y=10 then x-y= (a) 3 (b) 4 (c) 5 (d) 6

If x+y=k is normal to y^2=12 x , then k is (a)3 (b) 9 (c) -9 (d) -3

If x+y=k is normal to y^2=12 x , then k is (a)3 (b) 9 (c) -9 (d) -3

If x+y=k is normal to y^2=12 x , then k is (a)3 (b)9 (c) -9 (d) -3

If x+y=k is normal to y^2=12 x , then k is 3 (b) 9 (c) -9 (d) -3

If x^3-2x^2y^2+5x+y-5=0 and y(1)=1 , then (a) y^(prime)(1)=4/3 (b) y'(1)=-4/3 (c) y''(1)=-8(22)/(27) (d) y^(prime)(1)=2/3

If x^3-2x^2y^2+5x+y-5=0 and y(1)=1 , then (a) y^(prime)(1)=4/3 (b) y'(1)=-4/3 (c) y''(1)=-8(22)/(27) (d) y^(prime)(1)=2/3

If x^3-2x^2y^2+5x+y-5=0 and y(1)=1 , then (a) y^(prime)(1)=4/3 (b) y'(1)=-4/3 (c) y''(1)=-8(22)/(27) (d) y^(prime)(1)=2/3

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))

x ^ (2) + y ^ (2) + 1, x ^ (2) + 2y ^ (2) + 3, x ^ (2) + 3y ^ (2) + 4y ^ (2) + 2,2y ^ (2) + 6.3y ^ (2) + 8y ^ (2) + 1.2y ^ (2) + 3.3y ^ (2) +4] |