The correct option is A 2016
2x−y=12 and xy=4
8x3−y3=(2x)3−y3
=(2x−y)((2x)2+y2+2xy) [Using a3−b3=(a−b)(a2+b2+ab)]
=(2x−y)[(2x−y)2+2×2x×y+2xy]
[Using a2+b2=(a−b)2+2ab]
=(2x−y)[(2x−y)2+6xy]
On substituting the values, 2x−y=12 and xy=4, we get,
=12×[(12)2+6(4)]
=12[144+24]
=12×168=2016