Answer
$\dfrac{8y^4\sqrt{2m}}{m^2}$
Work Step by Step
Getting the factor that is a perfect root of the index, the given expression, $
\sqrt{\dfrac{2^7y^8}{m^3}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\sqrt{\dfrac{2^6y^8}{m^2}\cdot\dfrac{2}{m}}
\\\\=
\sqrt{\left(\dfrac{2^3y^4}{m}\right)^2\cdot\dfrac{2}{m}}
\\\\=
\dfrac{2^3y^4}{m}\sqrt{\dfrac{2}{m}}
.\end{array}
Rationalizing the denominator results to
\begin{array}{l}\require{cancel}
\dfrac{2^3y^4}{m}\sqrt{\dfrac{2}{m}\cdot\dfrac{m}{m}}
\\\\=
\dfrac{2^3y^4}{m}\sqrt{\dfrac{2m}{(m)^2}}
\\\\=
\dfrac{2^3y^4}{m}\cdot\dfrac{\sqrt{2m}}{\sqrt{(m)^2}}
\\\\=
\dfrac{2^3y^4}{m}\cdot\dfrac{\sqrt{2m}}{m}
\\\\=
\dfrac{2^3y^4\sqrt{2m}}{m^2}
\\\\=
\dfrac{8y^4\sqrt{2m}}{m^2}
.\end{array}
