Пусть b⁰,⁵=√b=t ⇒ b=t² b¹,⁵=(√b²)³=t³ ⇒
((t+3)/(t³-3t²)-(t-3)/(t³+3t²))*(t²-9)/t
1. (t+3)/(t²*(t-3))-(t-3)/(t²*(t+3))=((t+3)²-(t-3)²)/(t²*(t+3)*(t-3))=
=(t²+6t+9-t²+6t-9)/(t²*(t²-9))=12t/(t²*(t²-9)=12/(t*(t²-9)).
2. (12/(t*(t²-9))*(t²-9)/t=12*(t²-9)/(t*t*(t²-9))=12/t².
3.12/t²=12/(√b)²=12/b.
Ответ: 12/b.