$$A^{1\over A} = B^{1\over B} = C^{1\over 3} $$ and $$ A^{BC}+B^{AC}+C^{AB}=729 $$ Find the value of A^{1\A}.
What I tried doing - $$ A^{1\over A} = B^{1\over B} = C^{1\over 3} = k $$ $$ => A = k^A $$ $$ => B = k^B $$ $$ => C = k^3 $$
I tried substituting the values in the 2nd equation but that didn't lead me anywhere. $$A^{1\over A} = C^ {1\over 3} $$ $$ => A = C^{A\over3}$$ $$ => B = C^{B\over3}$$ $$ => AB = C^{A+B\over3}$$ Tried doing this as well but still didn't look right. Am I on the right track? Or am I completely wrong?
assuming that $$A^{1/A}=B^{1/B}=C^{1/C}=t$$ then we get using the second equation $$3t^{ABC}=729$$ therefore $$t^{ABC}=3^4$$ can you finish?