<s>[INST] Find the value of $\frac{1}{a^3}+\frac{1}{b^3}$ 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?
 [/INST] 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 :).
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