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Hint: In this question it is given that $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} },\ A^{BC}+B^{AC}+C^{AB}=729$$. We have to find the value of $$A^{\dfrac{1}{A} }$$. So to find the solution we first need to express A in terms of B and similarly express C in terms of B and we have to put in the second equation from where we will get the value of B and after that we can easily find the value of $$A^{\dfrac{1}{A} }$$.

__Complete step-by-step solution:__

Given,

$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$..................(1)

Taking first and second from the above equation, we get,

$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$

$$\Rightarrow \left( A^{\dfrac{1}{A} }\right)^{AB} =\left( B^{\dfrac{1}{B} }\right)^{AB} $$ [ taking power AB on the both side of the equation]

$$\Rightarrow \left( A\right)^{\dfrac{1}{A} \times AB} =\left( B\right)^{\dfrac{1}{B} \times AB} $$ [since, $$\left( a^{n}\right)^{m} =a^{n\times m}$$]

$$\Rightarrow A^{B}=B^{A}$$...........(2)

Now taking second and third form the equation(1) we get,

$$B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$

$$\Rightarrow \left( B^{\dfrac{1}{B} }\right)^{BC} =\left( C^{\dfrac{1}{C} }\right)^{BC} $$ [ taking Power BC on the both side]

$$\Rightarrow \left( B\right)^{\dfrac{1}{B} \times BC} =\left( C\right)^{\dfrac{1}{C} \times BC} $$

$$\Rightarrow B^{C}=C^{B}$$

$$\Rightarrow C^{B}=B^{C}$$...............(3)

Now another equation that is given,

$$A^{BC}+B^{AC}+C^{AB}=729$$

$$\Rightarrow \left( A^{B}\right)^{C} +B^{AC}+\left( C^{B}\right)^{A} =729$$.............(4)

Now putting the value of $$A^{B}\ \text{and} \ C^{B}$$ in the above equation we get,

$$ \left( B^{A}\right)^{C} +B^{AC}+\left( B^{C}\right)^{A} =729$$

$$\Rightarrow B^{AC}+B^{AC}+B^{AC}=729$$

$$\Rightarrow 3B^{AC}=729$$

$$\Rightarrow B^{AC}=\dfrac{729}{3}$$

$$\Rightarrow B^{AC}=243$$

$$\Rightarrow B^{AC}=3^{5}$$

So by comparing the sides we can say that the above equation is satisfied if and only if,

B=3 and AC=5

Now putting the value of B in equation (1) we get,

$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$

$$\Rightarrow A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$ [taking the first two]

$$\Rightarrow A^{\dfrac{1}{A} }=3^{\dfrac{{}1}{3} }$$ [since B=3]

Which is our required solution.

Note: You can also solve this problem by taking $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }=k$$, so from here you can able to find the value of A, B and C and after putting these values in the second equation you will get the value of k. Which helps you to find the value of $$A^{\dfrac{1}{A} }$$.

Given,

$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$..................(1)

Taking first and second from the above equation, we get,

$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$

$$\Rightarrow \left( A^{\dfrac{1}{A} }\right)^{AB} =\left( B^{\dfrac{1}{B} }\right)^{AB} $$ [ taking power AB on the both side of the equation]

$$\Rightarrow \left( A\right)^{\dfrac{1}{A} \times AB} =\left( B\right)^{\dfrac{1}{B} \times AB} $$ [since, $$\left( a^{n}\right)^{m} =a^{n\times m}$$]

$$\Rightarrow A^{B}=B^{A}$$...........(2)

Now taking second and third form the equation(1) we get,

$$B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$

$$\Rightarrow \left( B^{\dfrac{1}{B} }\right)^{BC} =\left( C^{\dfrac{1}{C} }\right)^{BC} $$ [ taking Power BC on the both side]

$$\Rightarrow \left( B\right)^{\dfrac{1}{B} \times BC} =\left( C\right)^{\dfrac{1}{C} \times BC} $$

$$\Rightarrow B^{C}=C^{B}$$

$$\Rightarrow C^{B}=B^{C}$$...............(3)

Now another equation that is given,

$$A^{BC}+B^{AC}+C^{AB}=729$$

$$\Rightarrow \left( A^{B}\right)^{C} +B^{AC}+\left( C^{B}\right)^{A} =729$$.............(4)

Now putting the value of $$A^{B}\ \text{and} \ C^{B}$$ in the above equation we get,

$$ \left( B^{A}\right)^{C} +B^{AC}+\left( B^{C}\right)^{A} =729$$

$$\Rightarrow B^{AC}+B^{AC}+B^{AC}=729$$

$$\Rightarrow 3B^{AC}=729$$

$$\Rightarrow B^{AC}=\dfrac{729}{3}$$

$$\Rightarrow B^{AC}=243$$

$$\Rightarrow B^{AC}=3^{5}$$

So by comparing the sides we can say that the above equation is satisfied if and only if,

B=3 and AC=5

Now putting the value of B in equation (1) we get,

$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$

$$\Rightarrow A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$ [taking the first two]

$$\Rightarrow A^{\dfrac{1}{A} }=3^{\dfrac{{}1}{3} }$$ [since B=3]

Which is our required solution.

Note: You can also solve this problem by taking $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }=k$$, so from here you can able to find the value of A, B and C and after putting these values in the second equation you will get the value of k. Which helps you to find the value of $$A^{\dfrac{1}{A} }$$.

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