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