Black box

Since we don't know how the resistors are connected, we can do some trial. We may see that:

$R_{AB} = \frac{5}{3} = 1 + \frac{2}{3} = 1 + \frac{1}{1 + \frac{1}{2}} = 1 + \frac{1}{1 + \frac{1}{\frac{1}{1} + \frac{1}{1}}}$

$R_{BC} = \frac{8}{3} = 1 + 1 + \frac{1}{1 + \frac{1}{\frac{1}{1} + \frac{1}{1}}}$

From above equation we can obtain this circuit. Hence, $R_{BC} = \frac{5}{3} \approx 1.6667$ $Ω$

With at most 5 resistors of $1 \Omega$ , there is only one way to create a circuit with equivalent resistance is $\frac{5}{3} \Omega$ and a circuit with $\frac{8}{3} \Omega$ , which show in this image .

Therefore, there is only one possible circuit for the circuit in the black box, which is also shown in this image .

Therefore, the equivalent resistance between B and C is $\frac{5}{3} \Omega$

For AB and AC same resistance there are two ways to connect . In one BC is $1\Omega$ . and in the other BC is $\dfrac 5 3 \Omega$ . as shown.

R {AB} = \frac{5}{3} >1 so we can obtain that there is one resistor connected in series with the other set of resistors. The value of that set is \frac{5}{3} -1 = \frac{2}{3}. It is clearly that this set contains (2 resistors in series, the joint of them is D) parallel with one resistor. R {AC}= \frac{8}{3} 1Ohm greater than R {AC}. Therefore, the possible arrangement is that D is connected with C by one resistor. So, R {BC} is ((R+R)//R)+R = \frac{5}{3}