remember
\(\frac{x^m}{x^n}=x^{m-n}\)
and
(ab)/c=(a/c)(b/c)
and
\(\sqrt[n]{x^m} =x^ \frac{m}{n}\)
and
\((x^m)^n=x^{mn}\)
and
x^-m=1/(x^m)
so
first convert 8 to 2³
\(\frac{ \sqrt[5]{2^3} \sqrt[2]{2^3} }{ \sqrt[3]{(2^3)^5} }\)=
\(\frac{(2^ \frac{3}{5})(2^ \frac{3}{2}) }{2^ \frac{15}{3} }\)=
we can simplify \(2^ \frac{15}{3}\) to \(2^5\)
\(\frac{(2^ \frac{3}{5})(2^ \frac{3}{2}) }{2^5}\)=
\(( \frac{2^ \frac{3}{5}}{2^5} )( \frac{2^ \frac{3}{2}}{2^5} )\)
(3/5)-5=3/5-25/5=-22/5
3/2-5=3/2-10/2=-7/2
\((2^ \frac{-22}{5})(2^ \frac{-7}{2})\)
-22/5+(-7/2)=
-44/10-35/10=
-79/10=-7 and 9/10
\(2^{-7&\frac{9}{10}=\frac{1}{7 \sqrt[10]{2^9} }\)
sorry for complex stuff
