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

Very nice! The only thing that I can add to this is that I would have dealt with $u^4+v^4$ in a way which is, I think, simpler than yours. Observe that\begin{align}u^4+v^4&=(u+v)^4-uv(4u^2+4v^2+6uv)\\&=(u+v)^4-uv\bigl(4(u+v)^2-2uv)\bigr)\\&=1\,296-uv(144-2uv)\\&=1\,296-144uv+2(uv)^2.\end{align}So, this leads to the equation$$2(uv)^2-144uv+1\,296=386,$$which is equivalent to$$(uv)^2-72uv+455=0.$$

$$\sqrt[4]{386-x}+\sqrt[4]{x}=6$$

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.

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.