<s>[INST] 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
 [/INST] 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.
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