<s>[INST] $A^ {1/A} =B^ {1/B} =C^ {1/C} ,A^ {BC} +B^ {AC} +C^ {AB} =729$ 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?
 [/INST] 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
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