Let $\square$ ABDC is cyclic quadrilateral , and $\triangle$ ABC is equilateral triangle of length a.
Express $ \overline {DA} ^4 + \overline {DB} ^4 + \overline {DC}^4 $ using terms of a.
Let $ \overline {DA} =x , \overline {DB} =y , \overline {DC} =z $
by Cosine law
$ x^2+ y^2 - xy = x^2 + z^2 -xz = y^2 +z^2 +yz = a^2 $
I think $ z \rightarrow 0 \ then\ y \rightarrow a ,\ x \rightarrow a $
so $x^4+y^4 +z^4 = 2a^4$ but it lacks persuation.
By P'tolemy, $ay + az = ax$
$ y+z = x $
By Cosine Law at $\triangle BDC$,
$ y^2 + z^2 - 2yz cos \frac {2 \pi} 3 = a^2 $
$ yz = x^2 -a^2 $
$y^2 + z^2 = 2a^2 -x^2 $
$ y^4 + z^4 = ( y^2 + z^2 )^2 - 2y^2 z^2 $
$ = (2a^2 -x^2 )^2 - 2( x^2 -a^2 )^2 $
$ = 2a^4 -x^4 $
so $ x^4 + y^4 + z^4 =2a ^4 $
After thinking all night, I was finally able to reach the answer. Thank you for your interest in this problem.