If x = (√2 + 1)/(√2 - 1) and y = (√2 - 1)/(√2 + 1), then 3 (xy + 1) is euqal to?

xy = [(√2 + 1)/(√2 - 1)] x [(√2 - 1)/(√2 + 1)] = 1
∴ 3(xy + 1) = 3(1 + 1) = 6
Now, x + y = [(√2 + 1)/(√2 - 1)] + [(√2 - 1)/(√2 + 1)]
= (√2 + 1)² + (√2 - 1)² / (√2 - 1) (√2 + 1)
= (2 + 1 +2√2) + (2 + 1 - 2√2) / (2 - 1) = 6
∴ 3(xy + 1) = x + y