Equivalent resistance of $R_{1}$ and $R_{2}$ (in parallel)
$R'=\frac{6 \times 6}{6+6}=3 \Omega$
Equivalent resistance of $R_{5}$ and $R_{7}$ (in parallel)
$R''=\frac{10 \times 10}{10+10}=5 \Omega$
Now, the circuit is reduced to
Now, this circuit can be shown as it is balanced
$\left(\frac{5}{3}=\frac{5}{3}\right)$
Total resistance between $A B$
$R_{A B} =\frac{(5+3)(5+3)}{(5+3)+(5+3)} $
$=\frac{8 \times 8}{8+8}=4\, \Omega$
