How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$?

I'm trying to guess a method for getting the values that work on this irrational equation: $$\sqrt[4]{x+27}+\sqrt[4]{55-x}=4, x\in\mathbb C$$

After using the formula $a^4+b^4=(a+b)(a^3-a^2b+ab^2+b^3)$ and doing some amplifications, I ended in this phase: $$p=x+27, r=55-x \\\sqrt[4]{p^3}+\sqrt[4]{r^3}-\sqrt{p}\sqrt[4]{r}+\sqrt[4]{p}\sqrt{r}=\frac{82}{4}$$ , which clearly is complicated than the initial equation.

Also, rising to the power of $4$, it isn't efficient either: you will end up with mixed radicals. Maybe I'm doing something wrong?

Let $\displaystyle x+27=a^4, 55-x=b^4\implies a^4+b^4=82$ and $a+b=4$

$\displaystyle a^4+b^4=(a^2+b^2)^2-2a^2b^2=\{(a+b)^2-2ab\}^2-2a^2b^2$ $\displaystyle=(16-2ab)^2-2a^2b^2=256+2a^2b^2-64ab$

$\displaystyle\implies 2(ab)^2-64ab+256=82\iff2(ab)^2-64ab+174=0$

Solve the quadratic equation for $ab$ and we already have $a+b=4$

Case $1: ab=3,$ this leads to too simple calculations

Case $2: ab=29,$ then $a,b$ are the roots of $\displaystyle t^2-4t+29=0$

$\displaystyle\implies (i)a,b $ are $2\pm5i$

and $\displaystyle(ii)t^2=4t-29$

$\displaystyle\implies t^4=(4t-29)^2=16t^2-29\cdot8t+29^2$ $\displaystyle=16(4t-29)-29\cdot8t+29^2=29(29-16)-8t(29-8)=377-168t$

$\displaystyle\implies$ the values of $\displaystyle a^4,b^4$ are $\displaystyle377-168(2\pm5i)=41\mp840i$

Find $x$ in either case and check if the values conform

Note that the function defined by the second radical is the mirror reflection of that defined by the first one, in the line $x=14$, with common range $-27\leqslant x\leqslant 55.$ Also, the fourth-root function has decreasing positive gradient. So the sum has decreasing positive gradient between $-27$ and $14$, zero gradient at $14$, and decreasing negative gradient between $14$ and $55$. At each end of the range, the sum is close to $3$, and it reaches its maximum of about $5$ at $x=14$. From this, it can be seen that the original equation has just two solutions. Now, if you look for an integer $x$ between $14$ and $55$ such that both $x+27$ and $55-x$ are fourth powers, you will easily find it; and symmetry will give you the other solution.

Suppose if u want x to be positive and integer then x=54 You know that second term 55 - x rays to 1/4 , if u make this term 1 which is equal to $1^4$ and first term x+27 to be $81=3^4$

so here x=54 is a trivialsolution while for method I agree with @lab bhattacharjee method