If $A^{1/A}=B^{1/B}=C^{1/C}$, $A^{BC}+B^{AC}+C^{AB}=768$. What is the value of $A^{1/A}$?

Question

If $A^{1/A}=B^{1/B}=C^{1/C}$, $A^{BC}+B^{AC}+C^{AB}=768$. What is the value of $A^{1/A}$?

I tried solving this question but each time I get different answers like $3^{1/3}$, 3, or 243. What would be the correct method to solve this question? Kindly elaborate your answer.The options are $81^{1/ABC}, 2^{1/2}, 27^{1/ABC}, 9^{1/ABC}$.I've changed the question as the previous question was incorrect.

Answer 1

$$\sum_{cyc}A^{BC}=\sum_{cyc}\left(A^{\frac{1}{A}}\right)^{ABC}=3\left(A^{\frac{1}{A}}\right)^{ABC},$$ which gives $$\left(A^{\frac{1}{A}}\right)^{ABC}=256,$$ which gives $$A^{BC}=B^{AC}=C^{AB}=256.$$ Also, since the equation $x^{\frac{1}{x}}=k$ has two roots maximum,

we obtain that at least two numbers from $\{A,B,C\}$ are equal.

Let $C=B.$

Thus, we got the following system: $$A^{B^2}=256$$ and $$B^{AB}=256.$$ We have $$A=256^{\frac{1}{B^2}},$$ which gives $$B^{B\cdot256^{\frac{1}{B^2}}}=256,$$ which gives $B=2$ or $B=1.029...$ or $B=2.474...$.

For $B=2$ we obtain: $$A^{\frac{1}{A}}=2^{\frac{1}{2}}=\sqrt2.$$