Answer
$\dfrac{(x+y)(x^2+xy+y^2)}{(x^2+y^2)}$
Work Step by Step
Multiplying by the reciprocal of the divisor, the given expression, $
\dfrac{3x+3y}{5x-5y}\div\dfrac{3x^2+3y^2}{5x^3-5y^3}
,$ is equivalent to
\begin{align*}
&
\dfrac{3x+3y}{5x-5y}\cdot\dfrac{5x^3-5y^3}{3x^2+3y^2}
.\end{align*}
Factoring the $GCF$ and cancelling the common factors, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{3(x+y)}{5(x-y)}\cdot\dfrac{5(x^3-y^3)}{3(x^2+y^2)}
\\\\&=
\dfrac{\cancel3(x+y)}{5(x-y)}\cdot\dfrac{5(x^3-y^3)}{\cancel3(x^2+y^2)}
\\\\&=
\dfrac{(x+y)}{\cancel5(x-y)}\cdot\dfrac{\cancel5(x^3-y^3)}{(x^2+y^2)}
\\\\&=
\dfrac{(x+y)}{(x-y)}\cdot\dfrac{(x^3-y^3)}{(x^2+y^2)}
.\end{align*}
Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ or the factoring of the difference of two cubes, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{(x+y)}{(x-y)}\cdot\dfrac{(x-y)(x^2+xy+y^2)}{(x^2+y^2)}
\\\\&=
\dfrac{(x+y)}{\cancel{(x-y)}}\cdot\dfrac{\cancel{(x-y)}(x^2+xy+y^2)}{(x^2+y^2)}
&(\text{cancel common factor})
\\\\&=
(x+y)\cdot\dfrac{(x^2+xy+y^2)}{(x^2+y^2)}
\\\\&=
\dfrac{(x+y)(x^2+xy+y^2)}{(x^2+y^2)}
.\end{align*}
Hence, the expression $\dfrac{3x+3y}{5x-5y}\div\dfrac{3x^2+3y^2}{5x^3-5y^3}$ simplifies to $\dfrac{(x+y)(x^2+xy+y^2)}{(x^2+y^2)}$.
