For the following circuit find the current through $6$ ohm and $3$ ohm.

I can solve with other method but please help me to know how to use symmetry methods to solve this.

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Imagine that the three $4\Omega$ resistors are removed, leaving two separate circuits.

Now join EF with a conducting wire so that they are at the same potential. No current will flow from one circuit into the other, because there is no path for it to return along, and charge which flows cannot accumulate in either of the two circuits.

How does the potential at C compare with that at D? How does the potential at A compare with that at B? The PD across each battery is divided equally across the resistors in the same circuit. Relative to the potentials at E=F as 0V, the potentials at A, B will be 2V and the potentials at C, D will be 1V.

What effect will it have if you re-insert resistors between AB, CD and EF? Since AB, CD and EF are all at the same potential, it will not make any difference if we connect AB, CD with wires or re-insert the $4\Omega$ resistors. No current will flow along these connecting wires/resistors.

So the currents in the $6\Omega$ and $3\Omega$ resistors are the same whether or not the $4\Omega$ resistors are there.

Why can't current enters from $EF$ and returns from $DC$ or $BA$, please help?

Such cases are prohibited by Kirchhoff's Laws.

For example, suppose current flows from A to B and back from F to E, in order to satisfy Kirchhoff's Current Law (KCL). Then the potential at A must be greater than that at B (because conventional current flows "downhill") and the potential at E must be less than that at F. This means that the potential drop across AE must be greater than across BF. However, applying Kirchhoff's Voltage Law (KVL) round the loops on the left and right, both potential drops across AE and BF must be 2V. So this situation cannot happen unless we have other voltage sources in the circuit.

Likewise if current flowed from A to B and also from E to F then we could satisfy KVL but not KCL. The potentials at A and E would be greater than at B and F respectively, while the potential drops across AE and BF are both 2V. However current would be flowing from left to right without any returning right to left. Current would disappear somewhere, So this cannot happen either, unless we have some current sources or sinks in the circuit.

For example, suppose current flows from A to B and back from F to E, in order to satisfy Kirchhoff's Current Law (KCL). Then the potential at A must be greater than that at B (because conventional current flows "downhill") and the potential at E must be less than that at F. This means that the potential drop across AE must be greater than across BF. However, applying Kirchhoff's Voltage Law (KVL) round the loops on the left and right, both potential drops across AE and BF must be 2V. So this situation cannot happen unless we have other voltage sources in the circuit.

Likewise if current flowed from A to B and also from E to F then we could satisfy KVL but not KCL. The potentials at A and E would be greater than at B and F respectively, while the potential drops across AE and BF are both 2V. However current would be flowing from left to right without any returning right to left. Current would disappear somewhere, So this cannot happen either, unless we have some current sources or sinks in the circuit.

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