$A^ {1/A} =B^ {1/B} =C^ {1/C} ,A^ {BC} +B^ {AC} +C^ {AB} =729$ If $A^
{1/A}
 =B^
{1/B}
 =C^ 
{1/C}
 ,A^
{BC}
 +B^
{AC}
 +C^
{AB}
 =729$
Which of the following equals $A^ 
{1/A}$?
I tried solving it and in the end got $A^{BC}= B^{AC}=C^{BA}= 3^5$ Thus $A^{1/A}=\ ^{ABC}\sqrt {3^5}$
Now I have two doubts. In some of the solutions I have seen, they have concluded that since $A^{BC}= 3^5, A= 3 \ and\ BC = 5$ Thus $ A^{1/A}= 3^{1/3}$. Is this even a correct method? Because using this logic, we can conclude the same that $B^{AC}= 3^5, thus, B= 3 \ and\ AC = 5$ and so does for $C$ making all three $A,B,C= 3 $ which contradicts the rest.
And my second doubt is the answer to my question however is given as $\sqrt2$. How does one get this answer?
 A: Let $A^{1/A} = B^{1/B} = C^{1/C} = k$. Now, raise each term to the $ABC$ power and you get:
$A^{BC} = B^{AC} = C{AB} = k^{ABC}$ We know that $A^{BC} + B^{AC} + C^{AB} = 729$.
So
it becomes,
$k^{ABC}\times k^{ABC}\times k^{ABC}=729$
$(k^{ABC})^3 = 729$
$k^{ABC} = 9$
$k = 9^{\frac{1}{ABC}}$
So, $A^{1/A} = 9^{1/ABC}$
Hope it helps
A: Although in the original question I see plus signs in the second equation, I am solving again the question, this time as if there were multiplication signs, as I have been suggested. The difference in the way of solving it is small anyway.
Let $A^{1/A}=B^{1/B}=C^{1/C}=K$ (Eq. 1).
Raising each term to the ABC power we get:
$A^{BC}=B^{CA}=C^{AB}=K^{ABC}$,
so the last equation of the question becomes:
$A^{BC}\times B^{CA}\times C^{AB}=729$ (Eq. 2),
$({K^{ABC}})^3=729=3^6$,
$K^{ABC}=3^2=9$ (Eq. 3).
On the other hand, the function $f(x)=x^{1/x}$ is defined for all x>0, is possitive for all x, and has one maximum at x=e, since, for $y=x^{1/x}$, we have:
$ln(y)=\frac{ln(x)}{x}$,
$\frac{y'}{y}=\frac{1-ln(x)}{x^2}$ and, since y and $x^2$ are always possitive:
y' is positive for ln(x)<1 (that is, f is an increasing function for x<e) and y' is negative for ln(x)>1 (that is, f is a decreasing function for x>e).
Since the limit of f(x) as x goes to infinity is 1, there are just two values of x with the same f(x) for $1<x<e^{1/e}$ and there is just on value of x with the same f(x) for $0<x\leq 1$ or $x=e^{1/e}$.
Thus there are two possibilities: A=B=C or there are two of them which are identical and the other is different.

*

*A=B=C: Then Eq 1 and 3 become:

$3^2=K^{A^3}=({A^{1/A}})^{A^3}=A^{\frac{A^3}{A}}=A^{A^{2}}$
and the same relation for B and for C, but I will not write it for simplicity. By taking logarithms in both sides of the equation and operating, we get:
$A^2*ln(A)=2*ln(3)$,
$2A^2*ln(A)=A^2*ln(A^2)=4*ln(3)$.
Introducing the variable $x=ln(A^2)$, the equation becomes:
$x*e^x=4*ln(3)$.
Remembering the Lambert W function defined as: $W(x*e^x)=x$, we get:
$W(4*ln(3))=x=ln(A^2)=2*ln(A)$, so:
$ln(A)=\frac{W(4*ln(3))}{2}$,
$A=B=C=e^{\frac{W(4*ln(3))}{2}}\approx 2.46661$. Then:
$K=(A)^{\frac{1}{A}}\approx 1.44199$.


*Two of them are equal and the other is different. For example, let us assume that $A=C\neq B$. Then Eq. 2 becomes:

$A^{AB}=3^2=9$, so $AB*ln(A)=2*ln(3)$, $B=\frac{2*ln(3)}{A*ln(A)}$ (Eq. 4),
$B^{A^2}=3^2=9$, so $A^2*ln(B)=2*ln(3)$, $B=e^{\frac{2*ln(3)}{A^2}}=3^{\frac{2}{A^2}}$ (Eq. 5).
When using together Eqs. 4 and 5, just one solution is obtained for A=C, being this solution the same as that found in case 1, so this case does not add any further solution, being the solution to this question unique.
