Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $

Question

$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

Answer 1

Let us try another way:

$xy=1$

$x+y=\dfrac{(3+\sqrt{5})^2+(3-\sqrt {5})^2}{9-5}=\dfrac{2(9+5)}{4}=7$

$x^3+y^3=(x+y)^3-3xy(x+y)=?$

Answer 2

You have made mistakes in computing $(3+\sqrt 5 )^{3}$ and $(3-\sqrt 5 )^{3}$. These are $72+32\sqrt 5$ and $72-32\sqrt 5$

Answer 3

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.

Answer 4

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:

$$322.$$