The equivalent resistance between the points and will be (each resistance is \(15 \Omega\) )

The circuit can be shown as given below

The equivalent resistance between and
\(\begin{aligned}R_{D C} &=\frac{15 \times(15+15)}{15+(15+15)}=\frac{15 \times 30}{15+30} \\&=\frac{15 \times 30}{45}=10 \Omega\end{aligned}\)
Now, between and the resistance of upper part
\(\mathrm{R}_{1}=15+10+15=40 \Omega\)
Between and, the resistance of middle part
\(\mathrm{R}_{2}=15+15=30 \Omega\)
Therefore, equivalent resistance between and
\(\begin{aligned}\frac{1}{\mathrm{R}^{\prime}} &=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15} \\&=\frac{3+4+8}{120}=\frac{15}{120} \Rightarrow \mathrm{R}^{\prime}=\frac{120}{15}=8 \Omega\end{aligned}\)