Real root demanded!

Let $a = \sqrt[4]{x+27}$ and $b = \sqrt[4]{55-x}$ . Then the equations become $a+b=4$ and $a^4 + b^4 = 82$ . So now we get $a^4 + (4-a)^4 = 82$ , which simplifies to $a^4-8a^3+48a^2-128a+87 = 0.$

The original equation has two solutions $a = 1$ and $a = 3$ by inspection, so the quartic polynomial above is divisible by $(a-1)(a-3)$ , and after some long division we get $(a-1)(a-3)(a^2-4a+29) = 0.$ The quadratic polynomial has no real roots, so the only solutions are $a = 1$ and $a = 3$ , or $x = -26$ and $x = 54$ , so the answer is $\fbox{28}$ .

First let's use the identity $(a+b)^4=a^4+b^4+4ab((a+b)^2-2ab)+6(ab)^2$ , that comes from expanding $(a+b)^4$ . Now, raise both sides to $4$ :

$(\sqrt[4]{x+27}+\sqrt[4]{55-x})^4=(4)^4 \\ x+27+55-x+4\sqrt[4]{(x+27)(55-x)}((\sqrt[4]{x+27}+\sqrt[4]{55-x})^2-2\sqrt[4]{(x+27)(55-x)})+6(\sqrt[4]{(x+27)(55-x)})^2=256 \\ 82+4\sqrt[4]{(x+27)(55-x)}((4)^2-2\sqrt[4]{(x+27)(55-x)})+6(\sqrt[4]{(x+27)(55-x)})^2=256$

Now let $t=\sqrt[4]{(x+27)(55-x)}=\sqrt[4]{-x^2+28x+1485}$ :

$82+4t(16-2t)+6t^2=256 \\ -174+64t-8t^2+6t^2=0 \\ -2t^2+64t-174=0 \\ t^2-32t+87=0 \\ (t-3)(t-29)=0 \\ t_1=3 \quad t_2=29$

Finally, undo the change, for $t_1$ we get:

$\sqrt[4]{-x^2+28x+1485}=3 \\ -x^2+28x+1485=81 \\ x^2-28x-1404=0 \\ (x-54)(x+26)=0 \\ x_1=54 \quad x_2=-26$

If we substitute these values on the original equation both of them work.

Undo the change for $t_2$ :

$\sqrt[4]{-x^2+28x+1485}=29 \\ -x^2+28x+1485=707281 \\ x^2-28x+707281=0 \\ x_3=14+180i \quad x_4=14-180i$

If we substitute these values, they don't work, also they aren't real, so the sum of all the real roots is $54-26=\boxed{28}$ .