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

I would simply let $x=u^4$ and hope that $386-u^4=(6-u)^4$ is easy to factor. Expanding and collecting first to
$$2u^4-24u^3+216u^2-864u+910=0$$
we can hope for a factorization 
$$u^4-12u^3+108u^2-432u+455=(u^2-au+b)(u^2-cu+d)$$
from which it would be easy to identify the real roots. (Hoping for an integer root would leave a cubic with a real root to solve for, so it's better to hope for a factorization into a pair of quadratics.)
The factorization $u^4+1\equiv(u^2+1)^2$ mod $2$ tells us $a$ and $c$ are both even, and the factorization $u^4\equiv (u^2-1)(u^2+1)$ mod $3$ tells us $a$ and $c$ are multiples of $3$.  This leaves two possibilities:
$$u^4-12u^3+108u^2-432u+455=(u^2+b)(u^2-12u+d)$$
and
$$u^4-12u^3+108u^2-432u+455=(u^2-6u+b)(u^2-6u+d)$$
The second of these somehow looks more likely.  It's satisfied if $108=b+d+36$, $432=6(b+d)$, and $bd=455$. We're in luck: the requirements $108=b+d+36$ and $432=6(b+d)$ are in accord; both say $b+d=72$.  So $b$ and $d$ are roots of the quadratic
$$v^2-72v+455=(v-7)(v-65)$$
We thus have
$$u^4-12u^3+108u^2-432u+455=(u^2-6u+7)(u^2-6u+65)$$
and we're just about done.  The quadratic $u^2-6u+65$ has complex roots whose real and imaginary parts are both non-zero (so that their fourth powers are not non-negative real numbers), while the roots of $u^2-6u+7$ are $u=3\pm\sqrt2$.  Thus $x=(3+\sqrt2)^4$ and $x=(3-\sqrt2)^4$ are the real roots of the original equation.
