Find answer for this question

Dear Student ,

Please find below the solution to the asked query :

Let space straight A to the power of 1 over straight A end exponent space equals space straight B to the power of 1 over straight B end exponent space equals space straight C to the power of 1 over straight C end exponent space equals straight k Then comma straight A equals straight k to the power of straight A space comma space straight B equals straight k to the power of straight B space comma space straight C equals straight k to the power of straight C Now comma straight A to the power of BC space plus space straight B to the power of AC space plus space straight C to the power of AB space equals space 729 open parentheses straight k to the power of straight A close parentheses to the power of BC space plus space open parentheses straight k to the power of straight B close parentheses to the power of AC space plus space open parentheses straight k to the power of straight C close parentheses to the power of AB space equals space 729 straight k to the power of ABC space plus straight k to the power of ABC space plus space straight k to the power of ABC space equals space 729 3 space. space straight k to the power of ABC space equals space 729 straight k to the power of ABC space equals space 243 straight k space equals open parentheses space 243 space close parentheses to the power of 1 over ABC end exponent straight A to the power of 1 over straight A end exponent equals open parentheses space 243 space close parentheses to the power of 1 over ABC end exponent

Hope this information will clear your doubts about the topic .

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible .

Regards

  • 0
What are you looking for?