Answer
The required solution is
$\frac{{{a}^{2}}+{{b}^{2}}}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}$.
Work Step by Step
We have the given algebraic expression
$\frac{ab}{{{a}^{2}}+ab+{{b}^{2}}}+\left( \frac{ac-ad-bc+bd}{ac-ad+bc-bd}\div \frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{3}}+{{b}^{3}}} \right)$.
Now, solve the bracket of the given expression:
$\frac{ac-ad-bc+bd}{ac-ad+bc-bd}\div \frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{3}}+{{b}^{3}}}$.
And the above expression can be written as
$\frac{ac-ad-bc+bd}{ac-ad+bc-bd}\div \frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{3}}+{{b}^{3}}}=\frac{ac-ad-bc+bd}{ac-ad+bc-bd}\times \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}$.
Now, factorize $ac-ad-bc+bd$ by taking out common terms:
$\begin{align}
& ac-ad-bc+bd=a\left( c-d \right)-b\left( c-d \right) \\
& =\left( c-d \right)\left( a-b \right)
\end{align}$.
Also, factorize $ac-ad+bc-bd$ by taking out common terms:
$\begin{align}
& ac-ad+bc-bd=a\left( c-d \right)+b\left( c-d \right) \\
& =\left( c-d \right)\left( a+b \right)
\end{align}$.
Further factorize ${{a}^{3}}-{{b}^{3}}$ by using the identity ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$:
$\begin{align}
& {{a}^{3}}-{{b}^{3}}={{\left( a \right)}^{3}}-{{\left( b \right)}^{3}} \\
& =\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)
\end{align}$.
Further factorize ${{a}^{3}}+{{b}^{3}}$ by using the identity ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$:
$\begin{align}
& {{a}^{3}}+{{b}^{3}}={{\left( a \right)}^{3}}+{{\left( b \right)}^{3}} \\
& =\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)
\end{align}$.
So, the bracket of the given expression gets reduced to
$\begin{align}
& \frac{ac-ad-bc+bd}{ac-ad+bc-bd}\div \frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{3}}+{{b}^{3}}}=\frac{\left( c-d \right)\left( a-b \right)}{\left( c-d \right)\left( a+b \right)}\times \frac{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)} \\
& =\frac{{{a}^{2}}-ab+{{b}^{2}}}{{{a}^{2}}+ab+{{b}^{2}}}
\end{align}$.
And simplify the given algebraic expression:
$\begin{align}
& \frac{ab}{{{a}^{2}}+ab+{{b}^{2}}}+\left( \frac{ac-ad-bc+bd}{ac-ad+bc-bd}\div \frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{3}}+{{b}^{3}}} \right)=\frac{ab}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}+\frac{{{a}^{2}}-ab+{{b}^{2}}}{{{a}^{2}}+ab+{{b}^{2}}} \\
& =\frac{ab+{{a}^{2}}-ab+{{b}^{2}}}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)} \\
& =\frac{{{a}^{2}}+{{b}^{2}}}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}
\end{align}$.
Hence, $\frac{ab}{{{a}^{2}}+ab+{{b}^{2}}}+\left( \frac{ac-ad-bc+bd}{ac-ad+bc-bd}\div \frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{3}}+{{b}^{3}}} \right)=$ $\frac{{{a}^{2}}+{{b}^{2}}}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}$.
