Question

if A^1/a=B^1/b=c^1/c and A^bc+B^ca+c^ab =729 then find B^1/b value

Answer 1

hey friend!!!!!
_______


Let

A^1/A=B^1/B=C^1/C=k

Then

,A=k^A. B=k^B. C=k^C

Now,

A^(B+C)+B(A+C)+C(A+B)=729

K^ABC+K^ABC+K^ABC=729


3×K^ABC=729

K^ABC=243

K^ABC=3^5

Comparing both sides

K=B^1/B=3___________answer.

hope it will help you..

Answer 2

$B^{1/B} = 9^{1/ABC}$

Step-by-step explanation:

Correction in Question

$if A^{1/A} = B^{1/B} = C^{1/C} , A^{BC} \times B^{AC} \times C^{AB} = 729$

Let say

$A^{1/A} = B^{1/B} = C^{1/C} = k$

Taking power ABC each side then

$if A^{BC} = B^{AC} = C^{AB} =K^{ABC}$

$A^{BC} \times B^{AC} \times C^{AB} = 729\\k^{ABC} \times k^{ABC} \times k^{ABC} = 729\\( k^{ABC})^3 = 729\\( k^{ABC})^3 = 9^3\\ k^{ABC} = 9$

$k = 9^{1/ABC}\\A^{1/A} = 9^{1/ABC}$

$B^{1/B} = 9^{1/ABC}$

Hence

$B^{1/B} = 9^{1/ABC}$

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