Find the resistance between point C and point B of the network as shown below :

Explanation

(a) For the net resistance between A and B, equivalent circuit can be drawn as :
Now `R_3 and R_4` are in series and let their net resistance be `R_(34)`
Now `R_(34) = R_2 + R_4= 6 + 3 = 9Omega`
Similarly `R_(25) = R_2 + R_5 = 8 + 6 = 14Omega`
Now `R_(34) and R_(25)`are in parallel and let their net resistance be `R_P`

`1/R_p=1/R_(34)+1/R_(25)`
`1/R_P =1/9 +1/14 = (14+9)/(14xx9) = 23/126`
`R_P = 126/23=5.48Omega`
Now `R_1, R_P and R_6` are in series and let their net resistance R is given by
`R= R_1 + R_P + R_6`
`R= 7 + 5.48 + 4`
`implies R= 16.48 Omega`
(b) For the net resistance between points C and D, equivalent circuit can be drawn as :

Here `R_2, R_3 and R_4` are in series and let their net resistance be `R_(234)`
`R_(234) = R_2 + R_3 + R_4`
`R_(234) = 6+3 +4 = 13Omega`
Now `R_(234) R_5` are in parallel and let their net resistance be `R_P`
`1/R_P=1/R_(234)+1/R_5=1/13+1/8`
`1/R_P=(8+13)/(13xx8)=21/104`
`R_P = 104/21=4.95Omega`
Now `R_1, R_p and R_6` are in series and let their net resistance be R.
So `R=R_(1)+R_P +R_(6)`
`= 7+4.95 + 9`
`implies R = 20.95 = 21Omega`