Answer
$\dfrac{3a}{5(a-b)}$
Work Step by Step
Factoring the expressions and then cancelling common factors between the numerator and the denominator, then the given expression, $
\dfrac{a^3+a^2b+a+b}{5a^3+5a}\cdot\dfrac{6a^2}{2a^2-2b^2}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{(a^3+a^2b)+(a+b)}{5a(a^2+1)}\cdot\dfrac{6a^2}{2(a^2-b^2)}
\\\\=
\dfrac{a^2(a+b)+(a+b)}{5a(a^2+1)}\cdot\dfrac{2\cdot3\cdot a\cdot a}{2(a-b)(a+b)}
\\\\=
\dfrac{(\cancel{a+b})(\cancel{a^2+1})}{5\cancel{a}(\cancel{a^2+1})}\cdot\dfrac{\cancel{2}\cdot3\cdot \cancel{a}\cdot a}{\cancel{2}(a-b)(\cancel{a+b})}
\\\\=
\dfrac{3a}{5(a-b)}
.\end{array}
