Question

please solve question no....58. plz give solution detail ।।।।spams will be reported to brainly​

Answer 1

$let \: \: \: \: a {}^{ \frac{1}{a} } = b {}^{ \frac{1}{b} } = c {}^{ \frac{1}{c} } = k \\ \\ a = k {}^{a} \: \: \: b = k {}^{b} \: \: \: \: and \: \: \: \: c = k {}^{c} \\ \\ \\ a {}^{bc} + b {}^{ac} + c {}^{ab} = 729 \: \: \: \: \: given \\ \\ \\ k {}^{abc} + k {}^{abc} + k{}^{abc} = 729 \\ \\ 3k {}^{abc} = 729 \\ \\ k {}^{abc} = \frac{729}{3} \\ \\ k {}^{abc} = 243 \\ \\ k = \sqrt[abc]{243} \\ \\ as \: \: \: we \: \: \: already \: \: \: know \: \: \: \: k = a {}^{ \frac{1}{a} } \\ \\ a {}^{ \frac{1}{a} } = \sqrt[abc]{243}$

Answer 2

$\huge\underline\frak{Correct\:Question :}$

If $\sf A^\frac{1}{A}=B^\frac{1}{B}=C^\frac{1}{C}$, $\sf A^{BC}\times B^{AC}\times C^{AB}=729$. Which of the following equals $\sf A^\frac{1}{A}$

$\huge\underline\frak{Answer :}$

$\bullet\:\:\sf{A^\frac{1}{A}=B^\frac{1}{B}=C^\frac{1}{C}}\\\bullet\:\:\sf{A^{BC}\times B^{AC}\times C^{AB}=729}\\\bullet\:\:\sf{A^\frac{1}{A}= \:?}$

Let, $\sf A^\frac{1}{A}=B^\frac{1}{B}=C^\frac{1}{C}=n$

$\rule{120}{1}$

Taking Power ABC each side.

$:\implies\tt A^\frac{1}{A}=B^\frac{1}{B}=C^\frac{1}{C}=n\\\\\\:\implies\tt A^{\frac{1}{A} \times ABC}=B^{\frac{1}{B} \times ABC}=C^{\frac{1}{C} \times ABC}=n^{ABC}\\\\\\:\implies\tt A^{BC}= B^{AC}= C^{AB}=n^{ABC}$

$\rule{200}{1}$

☯⠀$\underline{\textsf{We have Given that :}}$

$\dashrightarrow\tt\:\:A^{BC}\times B^{AC}\times C^{AB}=729\\\\\\\dashrightarrow\tt\:\:n^{ABC}\times n^{ABC}\times n^{ABC}=729\\\\\\\dashrightarrow\tt\:\:n^{{ABC}^3}=729\\\\\\\dashrightarrow\tt\:\:n^{ABC} = \sqrt[3]{729}\\\\\\\dashrightarrow\tt\:\:n^{ABC} = \sqrt[3]{9 \times 9 \times 9}\\\\\\\dashrightarrow\tt\:\:n^{ABC} = 9\\\\\\\dashrightarrow\tt\:\:\Large\underline{\boxed{\green{\tt n = \sqrt[ABC]{9} = A^\frac{1}{A}}}}$

⠀

$\therefore\:\underline{\sf{Correct \:Option \:will \:be\:\:d) \bf{\sqrt[ABC]{9}}}.}$