Resistors - 2

The two 3 $\Omega$ resistors in the top right corner are in series therefore total resistance of the two is 3+3=6.

These resistors are also in parallel with the 6 $\Omega$ resistor in the top right corner of the rhombus like arrangement of the resistors, therefore the total resistance will be- 1/(1/6+1/6)=1/(2/6)=6/2=3 $\Omega$ .

This 3 $\Omega$ is in series with the 3 $\Omega$ in the bottom right corner and total resistance will be=3+3=6.

This 6 $\Omega$ will be in parallel with the 6 $\Omega$ in the center and as we have seen above the total resistance will be 3 $\Omega$ .

This 3 $\Omega$ is in series with the 3 $\Omega$ in the bottom left corner so the total resistance of the two will be 3+3=6 $\Omega$ .

This 6 $\Omega$ is in parallel with the 6 $\Omega$ in the top left corner and thus the total resistance will be equal to 3 $\Omega$ .

The connection though looks complicated, but it is very easy,

Let's start from $A$ , we have two $3\Omega$ resistors in series, (let this be $R_1$ ), which is connected to a $6\Omega$ resistor in parallel (let this be $R_2$ ), therefore, the equivalent resistance is given by $\dfrac{1}{R_{eq}}= \dfrac{1}{R_1}+\dfrac{1}{R_2} \\ = \dfrac{1}{3+3}+\dfrac{1}{6} \\ =\dfrac{2}{6} \\ Or, R_{eq}=3\Omega$

Similarly, we find that the remaining give rise to the same pattern, therefore, the total resistance of the conductor, i.e $R_{total}=3\Omega$

Start form th point from where u don't have any branching u will solve it mentally I bet.

( ( ( ( 3 - 3 ) || 6 ) - 3 ) || 6 ) - 3 ) || 6 )

from uppermost right corner, 3+3=6, 1/(1/6+1/6)=3, 3+3=6, 1/(1/6+1/6)=3, 3+3=6, 1/(1/6+1/6)=3

In the rightmost top corner, the two 3 ohm resistors are connected in series with each other and their combination ( 3+3=6 ) is connected in parallel with the 6 ohm resistor. Hence the total resistance of the right top corner is 6 || 6 = 3 ohm. This 3 ohm is connected in series with the lowermost right 3 ohm resistor giving effective resistance of 3 + 3 = 6 ohm. This 6 ohm is again connected with the central 6 ohm resistor in parallel giving resistance 6||6=3 ohm. This 3 ohm is now in series with the lowermost left 3 ohm giving total as 3+3=6 ohm. Now this 6 ohm is connected in parallel with the 6 ohm resistor between points A and B giving total effective resistance between A and B as 6||6=3 ohm... That's it...

3+3=6,6||6=3,3+3=6,6||6=3,3+3=6,6||6=3