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.

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.