Label the inersecting point as C.
Now please identify that we have, the upper 3Ω and lower 6Ω resistance between A and B. Hence they are in parallel.
So their equivalent = 6x3/9 = 2Ω .....(1)
Between A and C we have 8Ω and 2Ω in parallel (their equivalent is, 8x2/10 = 1.6Ω) and between C and B we have 4Ω and 6Ω in parallel (their equivalent is, 4x6/10 = 2.4Ω), and then resistances between A and C are in series with those from C to B (their equivalent is, 1.6+2.4 = 4Ω).
This 4Ω is between A and B, but above 2Ω marked as (1) is also between A and B,
Therefore equivalent resistance between A and B is, 4x2/6 = 8/6 = 4/3 Ω.