TITLE: Find the real roots for $\displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.$
QUESTION [10 upvotes]: Find the real roots for $\displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.$

Question from a Math Olympiad (ES, 2005). Answer: $(3\pm \sqrt{2})^4$. 
My attempt: I will make my attempt below, but I think the approach might be too complicated... are there simpler approaches?
I started making the substitution 
$$u=\sqrt[4]{386-x},~~v=\sqrt[4]{x}$$
so that $u^4+v^4=386$. 
Therefore we can define a system of 2 equations, in order to hopefully find $u$ and $v$:
$$\left\{ 
\begin{array}{l}
u+v=6 \\ 
u^4+v^4=386\\ 
\end{array}
\right.
$$ 
If we use $p_k=u^k+v^k$, $e_1=(u+v)$ and $e_2=uv$, by Newton-Girard identities we get
$$
\left\{ 
\begin{array}{l}
p_4=e_1p_3-e_2p_2\\
p_3=e_1p_2-e_2p_1\\
p_2=e_1p_1-e_2\times 2\\
\end{array}
\right.
$$ 
But as we know that $p_1=e_1=u+v=6$, we can make the substitutions backwards, 
so that 
$$
\left\{ 
\begin{array}{l}
p_2=36-2 uv\\
p_3=6(36-2uv)-6uv=216-18uv\\
p_4=6(216-18uv)-uv(36-2 uv)\\
\end{array}
\right.
$$ 
But, as we know that $p^4=u^4+v^4=386$, so making the substitution $z=uv$ and some simplifications the last equation is equivalent to
$$-z^2+72z-455=0.$$
Solving this last equation we find the roots $7$ and $65$, therefore $uv$ will potentially have these two values. Now as we know that $u+v=6$ we can set a system to solve for $u$ and $v$:
$$
\left\{ 
\begin{array}{l}
u+v=6\\
uv=7~~\text{or}~~65\\
\end{array}
\right.
$$ 
That can be solved noticing the $u$ and $v$ are the roots of $P(z)=z^2-(u+v)z+uv$. When $uv=65$, $P(z)$ has only complex roots. When $uv=7$, $P(z)$ has 2 real roots. 
When using $uv=7$ the roots $(u,v)$ for $P(z)$ are, by the symmetry, either
$$(3-\sqrt{2},3+\sqrt{2})~~\text{or}~~(3+\sqrt{2},3-\sqrt{2}).$$
But as $x=v^4$ the final solution would be 
$$x=(3\pm \sqrt{2})^4=193\pm 132\sqrt{2}.$$
Both solutions comply with the initial restritions for the argument of the radicals: $0\le x\le 386$.
Questions: (a) is the development correct? at least the final answer matches Wolphram Alpha; (b) are there other approaches? other solutions are welcomed.
Sorry if this is a duplicate.

REPLY [6 votes]: $$\sqrt[4]{386-x}+\sqrt[4]{x}=6$$
With a hope to solve the equation with only one unknown remained, notice both terms on the left sum up to a constant, we may express them as the distance from the half of $6$, which is $3$. 
Let $\sqrt[4]{386-x}=3-t,\sqrt[4]{x}=3+t$
$$(t-3)^4+(t+3)^4=386\tag1$$
$$t^4+54t^2-112=0$$
$$(t^2+27)^2=112+27^2$$
Since we are seeking for real roots
$$t^2=-27+\sqrt{112+27^2}=2\implies t=\pm\sqrt2$$
So
$$\sqrt[4]{x}=3\pm\sqrt2\implies x=(3\pm\sqrt2)^4$$
*As seen in $(1)$, such substitution allows cancellation of terms when both expression add up.