Answer
$\dfrac{5(3a+2)}{a}$
Work Step by Step
Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $
\dfrac{5a^2-20}{3a^2-12a}\div\dfrac{a^3+2a^2}{2a^2-8a}\cdot\dfrac{9a^3+6a^2}{2a^2-4a}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{5a^2-20}{3a^2-12a}\cdot\dfrac{2a^2-8a}{a^3+2a^2}\cdot\dfrac{9a^3+6a^2}{2a^2-4a}
\\\\
\dfrac{5(a^2-4)}{3a(a-4)}\cdot\dfrac{2a(a-4)}{a^2(a+2)}\cdot\dfrac{3a^2(3a+2)}{2a(a-2)}
\\\\
\dfrac{5(a+2)(a-2)}{3a(a-4)}\cdot\dfrac{2a(a-4)}{a^2(a+2)}\cdot\dfrac{3a^2(3a+2)}{2a(a-2)}
\\\\
\dfrac{5(\cancel{a+2})(\cancel{a-2})}{\cancel{3}a(\cancel{a-4})}\cdot\dfrac{\cancel{2a}(\cancel{a-4})}{\cancel{a^2}(\cancel{a+2})}\cdot\dfrac{\cancel{3}\cancel{a^2}(3a+2)}{\cancel{2a}(\cancel{a-2})}
\\\\=
\dfrac{5(3a+2)}{a}
.\end{array}
