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This image illustrates the step-by-step simplification of a complex resistor network into an equivalent single resistance between points A and B. The goal is to find the total resistance of the original circuit shown in part (a).

Let's break down the process shown in the diagram:

Step 1: Analyze the Original Circuit (Part a)

The original circuit has six resistors, each with a value of 3 Ω, arranged in a specific configuration between terminals A and B.

- Resistors R2 and R3 are connected in series between the top-left node and point C.
- Resistors R4 and R5 are connected in series between the bottom-left node and point D.
- Resistors R1 and R6 are connected directly from points C and D to terminals A and B, respectively.
- There is a direct connection (a wire) between points C and D, which means they are at the same electrical potential.

Step 2: Simplify the Left Side (Part b)

The first simplification combines the series resistors on the left side of the circuit.

- R2 and R3 are in series. The equivalent resistance for two resistors in series is the sum of their resistances.
- $R_{eq} = R_2 + R_3 = 3\ \Omega + 3\ \Omega = 6\ \Omega$
- Similarly, R4 and R5 are in series.
- $R_{eq} = R_4 + R_5 = 3\ \Omega + 3\ \Omega = 6\ \Omega$

However, looking at part (b), it appears that the diagram has combined these two 6 Ω branches into a single 9 Ω resistor labeled R5. This suggests a different interpretation or a typo in the labeling. Let's follow the logic presented in the diagram.

In part (b), the two 6 Ω branches (from R2+R3 and R4+R5) are shown as being connected in parallel between the input node and the common node CD. The equivalent resistance of two 6 Ω resistors in parallel is:
$R_p = \frac{1}{\frac{1}{6} + \frac{1}{6}} = \frac{1}{\frac{2}{6}} = \frac{6}{2} = 3\ \Omega$

But the diagram shows a 9 Ω resistor. This is inconsistent with standard circuit theory. It's possible the diagram intends to show the combination of R2, R3, R4, and R5 as a single equivalent resistance, but the value of 9 Ω does not match the calculation. Let's assume there might be an error in the diagram's labeling or value for this step and proceed with the next simplification.

Step 3: Further Simplification (Part c)

In part (c), the circuit is further simplified. The 9 Ω resistor from part (b) is replaced by a 2.25 Ω resistor labeled Rp. This suggests that the 9 Ω resistor was meant to represent a different combination or there's another layer of simplification not fully explained.

Given the final result in part (d), let's work backwards to understand the intended simplification.

Step 4: Final Equivalent Circuit (Part d)

Part (d) shows the final simplified circuit as three resistors in series between A and B:
- R1 = 3 Ω
- Rp = 2.25 Ω
- R6 = 3 Ω

The total resistance is the sum of these series resistors:
$R_{total} = R_1 + R_p + R_6 = 3\ \Omega + 2.25\ \Omega + 3\ \Omega = 8.25\ \Omega$

Conclusion

The problem is solved by systematically simplifying the resistor network. The key steps involve identifying series and parallel combinations and replacing them with their equivalent resistances.

The final answer is that the total resistance between points A and B is 8.25 Ω.

Note: There appears to be an inconsistency in the intermediate steps (part b and c) regarding the values of the equivalent resistors. However, the final result in part (d) is consistent with the expected outcome of the simplification process.