Solution set of the equation ${(x-1)}^4+{(x-5)}^4=82$  is

Given equation is ${(x-1)}^4+{(x-5)}^4=82$

Now splitting $-1$  as $-3+2$ , $-5$  as $-2-3$

$\Rightarrow {(x-3+2)}^4+{(x-3-2)}^4=82$

Let $x-3=y$

$\Rightarrow {(y+2)}^4+{(y-2)}^4=82$

$\Rightarrow \left[{(y+2)}^2{(y+2)}^2\right]+\left[{(y-2)}^2{(y-2)}^2\right]=82$

Expand and simplify we get,

$\Rightarrow 2y^4+48y^2+32=82$

$\Rightarrow 2\left(y^4+24y^2+16\right)=82$

$\Rightarrow y^4+24y^2-25=0$

$\Rightarrow y^2=-25,1$

$\Rightarrow y=\pm 5i,\pm 1$

$\Rightarrow x-3=\pm 5i,\pm 1$    $\left(\because x=y-3\right)$

$\Rightarrow x=\pm 5i+3,x=\pm \text{1+3}$

$\therefore$  The solution set of given equation is $\left\{2,4,3\pm 5i\right\}.$