Explanation

Put t = x+1/x: t³−3t = 65/8 gives 8t³−24t−65 = 0, with root t = 5/2, so x ∈ {2, 1/2}. Put s = y+1/y: 27s³−81s−730 = 0 has root s = 10/3, so y ∈ {3, 1/3}. Then xy ∈ {6, 3/2, 2/3, 1/6}; among the options only 6 occurs — option B.