Math, That looks not so unlike your cube problem, earlier, once it has been simplified. It has a not-dissimilar symmetry, too. (The bottom three resistors are a mirror image of the top three.)
But the answer is, of course, well expressed in Hamlet: "There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy." (Act 1, Scene 5.)
In short, there are many ways to approach your specific problem.
Note:
If B is grounded:
Then it must be the case that \$V_\text{D}+V_\text{E}=V_\text{C}+V_\text{F}=V_\text{A}\$ and we can replace \$V_\text{E}\$ and \$V_\text{F}\$, as shown above. That's promising as it's fewer unknowns.
Also, since two of those nodes are voltage sources, we can further reduce the above schematic:
This will immediately eliminate both of the remaining variables: \$V_\text{C}=\frac{3}{4}V_\text{A}\$ and \$V_\text{D}=\frac{5}{8}V_\text{A}\$:
$$\require{cancel}\begin{align*}\frac{V_\text{A}-V_\text{C}}{1\:\Omega}+\frac{V_\text{A}\cancel{-V_\text{D}}-V_\text{C}}{2\:\Omega}+\frac{\cancel{V_\text{D}}-V_\text{C}}{2\:\Omega}&=0\\\\\frac{3}{2}V_\text{A}-2V_\text{C}&=0\\\\V_\text{C}&=\frac34V_\text{A}\\\\\frac{V_\text{A}-V_\text{D}}{3\:\Omega}+\frac{V_\text{A}\cancel{-V_\text{C}}-V_\text{D}}{2\:\Omega}+\frac{\cancel{V_\text{C}}-V_\text{D}}{2\:\Omega}&=0\\\\\frac{5}{6}V_\text{A}-\frac43V_\text{D}&=0\\\\V_\text{D}&=\frac58V_\text{A}\end{align*}$$
And so, looking at the sum of the two currents going from \$V_\text{A}\$ to each of the those two following nodes, and dividing that into \$V_\text{A}\$, the equivalent resistance must be \$\frac83\:\Omega\$.
There are other ways. But that's just one more to add to the pile.