<s>[INST] Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
 a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be  $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $ 
I made this far: 
$$ 
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
 [/INST] From the numerator and denominator take $-n^{3}$ as common term and then rationalize the numerator again. 
Once you do that you will be left with :
$$\frac{n^{3}}{(n^{2}+\sqrt{n^{4}+1})(\sqrt{n^{2}+\sqrt{n^{4}+1}}+n\sqrt{2})}$$
Divide the numerator and denominator by $n^{3}$, and once you divide the denominator by $n^{3}$ for the first factor divide n^{2} and the second factor just n. You will get :
$$\frac{1}{(1+\sqrt{(1/n^{4})+1})(\sqrt{1+\sqrt{(1/n^{4})+1}}+\sqrt{2})}$$
Apply the limit you will get the answer as $\frac{1}{4\sqrt{2}}$
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