<s>[INST] Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$ 
Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$

This is my textbook contest exercise.
$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$
$$x=\big(\sqrt [4]{2-x}+1\big)^4$$
I know that
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
So, we will have some cube powers which introduce still $\sqrt [4]{\cdot}$ roots.
This makes more complicated the equation. How can I find the minimal polynomial using a more simple way?
 [/INST] Let $x=u^4,\;2-x=v^4$ with the restriction $u,v≥0$, then we have:
$$\begin{align}&\begin{cases}u-v=1\\u^4+v^4=2\end{cases}
\\\implies &v^4+(v+1)^4=2\end{align}$$
Then, using the key substitution $v+\frac 12=y$ leads to:
$$\begin{align}&\left(y-\frac 12\right)^4+\left(y+\frac 12\right)^4=2\\
\implies &16y^4+24y^2-15=0\end{align}$$
Finally, substitute $y^2=z$, you obtain:
$$16z^2+24z-15=0$$
The solution is completed by reversing the steps.
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