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

Someone has already pointed out where your mistake is. This is just to point to another alternative, for the fun of it.
You can factor $x^3+y^3$ as $$(x+y)(x^2-xy+y^2).$$ To compute the sum of squares, note that $2(x^2+y^2)=(x+y)^2+(x-y)^2.$ So, you only need to compute the sum, difference and product of your numbers, and combine these accordingly. This simplifies the calculation.
