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?
 A: 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
A: 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.    
A: 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
A: Setting
$x+27=u^{4}$
and
$55-x=v^{4}$,
we have:
$u+v=4$
and
$u^{4}+v^{4}=82$
The two numbers that meet these requirements are:
$u=1$ and $v=3$
therefore
$x=-26$ and $x=54$.
