The equivalent resistance is calculated as follows:
\[ R_{\text{eq}} = \frac{15 \times 10}{15 + 10} = \frac{150}{25} = 6 \, \Omega \]
Diode 2 is in reverse bias, so no current will flow through the branch containing it. It can be treated as a broken wire.
Diode 1 is in forward bias and will behave as a conducting wire. The new circuit becomes a simpler parallel and series resistor network.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: