<s>[INST] 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?
 [/INST] 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
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