Question

If $A^{\frac{1}{A}}=B^{\frac{1}{B}}=C^{\frac{1}{C}}$ and $A^{BC}+B^{CA}+C^{AB}=729$, then the value of $B^{\frac{1}{B}}$ is

• A. ${243}^{\frac{1}{ABC}}$
• B. ${81}^{\frac{1}{ABC}}$
• C. $3^{\frac{1}{ABC}}$
• D. $3^{ABC}$

Answer 1

The correct option is A ${243}^{\frac{1}{ABC}}$

Given: $A^{\frac{1}{A}}=B^{\frac{1}{B}}=C^{\frac{1}{C}}$ .....(i)
$A^{BC}+B^{CA}+C^{AB}=729$ ...(ii)
From (ii):
${\left(A^{\frac{1}{A}}\right)}^{ABC}+{\left(B^{\frac{1}{B}}\right)}^{ABC}+{\left(C^{\frac{1}{C}}\right)}^{ABC}=729$
$\Rightarrow 3{\left(B^{\frac{1}{B}}\right)}^{ABC}=729$ [From (i)]
$\Rightarrow {\left(B^{\frac{1}{B}}\right)}^{ABC}=243$
$\Rightarrow B^{\frac{1}{B}}={\left(243\right)}^{\frac{1}{ABC}}$
Hence, the correct answer is option a.