Possible values of A and B, keeping in mind that 30 < C(2) < 100, are:
[A = 1, B = 4; A^{3} + B^{3} = 65], [A = 2, B = 3; A^{3} + B^{3} = 35], [A = 2, B = 4; A^{3} + B^{3} = 72],
[A = 3, B = 3; A^{3} + B^{3} = 54], [A = 3, B = 4; A^{3} + B^{3} = 91], [A = 4, B = −1; A^{3} + B^{3} = 63],
[A = 4, B = −2; A^{3} + B^{3} = 56], [A = 4, B = −3; A^{3} + B^{3} = 37], [A = 5, B = −3; A^{3} + B^{3} = 98],
[A = 5, B = −4; A^{3} + B^{3} = 61], [A = 6, B = −5; A^{3} + B^{3} = 91]
We see that the only sum which occurs twice is 91, and hence that C(2) must be 91, since it is the smallest number in this list expressible as the sum of two cubes in two different ways.
∴ 91 = 3^{3} + 4^{3} = 6^{3} − 5^{3}
∴ ABXY = 3 × 4 × 5 × 6 = 360
Hence, option 1.
