Find potential difference between nodes A and B in circuit below

Something tells me this question is abandoned by the author, but I'll give my view anyway as it's a funny one. I assume $I_x=-I_z$ and A is only connected to the $8,Omega$ and the $6, V$ source, i.e.: crosses are no dots.
Then the 2 current sources are in parallel and their series resistors can be neglected, i.e.: there is a current source of $10 , A$ connected from C to D.
From KVL $8 , Omega I_x - 10 , V - 8 , Omega I_y - 6 , V = 0$ and from KCL $I_x + I_y = 10 , A $. In other words:

$$ I_x = 10 , A - I_y , , , I_x = 2 , A +I_y $$ From subtracting the second from the first we have $ 0 = 8 , A -2I_y $ so $$ I_y = 4, A , , , I_x = 6, A$$ making $$ U_A - U_B = 6 , V + 8 , Omega I_x = 54 , V $$