Question

# If $A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} },\ A^{BC}+B^{AC}+C^{AB}=729$. Find the value of $A^{\dfrac{1}{A} }$.

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 $A^{\dfrac{1}{A} }$.