An algebra problem by Wildan Bagus Wicaksono

Let $y=\sqrt [ 4 ]{ x }$

$y=\frac { 12 }{ 7-y } \\ y-\frac { 12 }{ 7-y } =0\\ \frac { y(7-y)-12 }{ 7-y } =0\\ \frac { 7y-{ y }^{ 2 }-12 }{ 7-y } =0\qquad \qquad \\ 7y-{ y }^{ 2 }-12=0\\ { y }^{ 2 }-7y+12=0\\ (y-3)(y-4)=0\\ { y }_{ 1 }=3\\ { y }_{ 2 }=4$

For $y = 3$ , obtained $\sqrt [ 4 ]{ x } =3\Rightarrow x=81$

For $y = 4$ , obtained $\sqrt [ 4 ]{ x } =4\Rightarrow x=256$

So, the solution of the equation is $81 + 256 = 337$ .