Find the value of $\frac{1}{a^3}+\frac{1}{b^3}$

Question

Let be a, b solutions for the equation $x^2-2\sqrt{\sqrt{2}+1} (x)+\sqrt{\sqrt{2}+1}=0$

Find the value of $\frac{1}{a^3}+\frac{1}{b^3}$

Using Vieta's formula $a+b=2c$ , $ab=c$ (where $c=\sqrt{\sqrt{2}+1}$) and solving I find that the answer is

$\frac{1}{a^3}+\frac{1}{b^3}=\frac{(a+b)(a^2-ab+b^2)}{(ab)^3}=8-\frac{6}{c}$. It's correct?

Answer 1

Let $ c = \sqrt{\sqrt{2} + 1} $. Then your given equation is written as :

$ x^2 - 2cx + c = 0 $

Solving it gives you : $ Δ = 4c^2 - 4c > 0 $

$ x_{1,2} = \frac {- 2c \pm \sqrt{Δ}}{2} $

You can then proceed with the calculations by substituting $c$ back in. Just be careful with the square roots and your calculations :).