thevenin resistance in cube
First of all, you can thevenize the 40ohm resistor and the 1A source. Then you have a 40V voltage source in series with a 40ohm resistor.
Now this 40 ohm resistor is in series with the 20 ohm one, so you have just a 60ohm resistor. You have reduced the circuit to a two-node, two-mesh one. You can go further and thevenise the 88V voltage source and the 10 ohm and the 50 ohm resistor to obtain a 73.333V source in series with an 8.3333 ohm resistor. The circuit now becomes just one mesh, with two voltage sources and one resistor, 68.333 ohm. Calculate the current through it: (73.333-40)/68.333=0.487A
The voltage drop across the 8.333 ohm resistor would be 8.333*0.487=4.065V. This gets subtracted from the 73.333V, to get the open-circuit voltage of 73.333-4.065=69.27V This is Voc.
For the short-circuit current, the 50ohm resistor is zero, so the 88V source will supply a current through the 10 ohm resistor: 88/10=8.8A.
The 1A source will supply 1A, which will be divided between the 20ohm and the 40ohm resistors. Through the 20ohm one you will have then two thirds of the 1A, that is 0.666A. Thus, the total SC current is: Isc= 8.8+0.666=9.466A
With that, Rth=69.27V/9.466A=7.317 ohm.
Does this result seem right? Let's see, if there was only the 88V source and the 10ohm resistor we would have just that, an 88V source and 10ohms of resistance. Since we have more resistance in parallel, the total resistance will be lower. Since the 10ohm is the lowest, it seems OK that the total resistance is not much lower than 10 ohms.
The voltage is 69.27V. That is close to, but lower than 88V, which seems OK, given the divider involving a 10ohm resistor and the rest; and the 1A times 40ohm is 40V, so if this was alone in the circuit, the final voltage would have been lower than 69.27V. So the 69.27V looks OK.