<s>[INST] Solve the equation with radicals: $x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0$ Solve this equation :
$ x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0 $
I've calculated $ \Delta $ and it is something like this: 
$ \sqrt[4]{4} + 4 \sqrt[4]{2} + 4 - 4 \sqrt[4]{8} $
I don't know what to do next.
 [/INST] Let us denote $a=\sqrt[4]2$. (Maybe this will help to notice how coefficients are related.)
We can rewrite the original equation as
\begin{align*}
x^2-xa(1+a)+a^3&=0\\
x^2-xa-xa^2+a^3&=0\\
x(x-a)-a^2(x-a)&=0\\
(x-a^2)(x-a)&=0
\end{align*}
This only has solutions $x_1=a$ and $x_2=a^2$. 
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