Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{47 + 32\sqrt5} = 2*7329/-2911$
why my answer is wrong? please help me

Numerator:
$$(3-\sqrt5)^6+(3+\sqrt5)^6=(14-6\sqrt5)^3+(14+6\sqrt5)^3=8\,(7-3\sqrt5)^3+8(7+3\sqrt5)^3\\
=16\,(7^3+3\cdot7\cdot9\cdot5)=16\cdot7\cdot184.$$
Denominator:
$$(3-\sqrt5)^3(3+\sqrt5)^3=4^3.$$
Ratio:
$$322.$$
