Answer
The required solution is:
$\frac{2d}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}$.
Work Step by Step
We have the given algebraic expression
$\left( \frac{1}{{{a}^{3}}-{{b}^{3}}}\cdot \frac{ac+ad-bc-bd}{1} \right)-\frac{c-d}{{{a}^{2}}+ab+{{b}^{2}}}$.
Solve the bracket of the given expression:
$\frac{1}{{{a}^{3}}-{{b}^{3}}}\cdot \frac{ac+ad-bc-bd}{1}$.
Now factorize ${{a}^{3}}-{{b}^{3}}$ by using the identity ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$:
$\begin{align}
& {{a}^{3}}-{{b}^{3}}={{\left( a \right)}^{3}}-{{\left( b \right)}^{3}} \\
& =\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)
\end{align}$.
Factorize $ac+ad-bc-bd$ by taking out common terms:
$\begin{align}
& ac+ad-bc-bd=a\left( c+d \right)-b\left( c+d \right) \\
& =\left( c+d \right)\left( a-b \right)
\end{align}$.
Therefore, the bracket of the given expression gets reduced to
$\begin{align}
& \frac{1}{{{a}^{3}}-{{b}^{3}}}\cdot \frac{ac+ad-bc-bd}{1}=\frac{1}{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)}\cdot \frac{\left( c+d \right)\left( a-b \right)}{1} \\
& =\frac{\left( c+d \right)}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}
\end{align}$.
Simplify the given algebraic expression:
$\begin{align}
& \left( \frac{1}{{{a}^{3}}-{{b}^{3}}}\cdot \frac{ac+ad-bc-bd}{1} \right)-\frac{c-d}{{{a}^{2}}+ab+{{b}^{2}}}=\frac{\left( c+d \right)}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}-\frac{c-d}{{{a}^{2}}+ab+{{b}^{2}}} \\
& =\frac{c+d-c+d}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)} \\
& =\frac{2d}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}
\end{align}$.
Therefore, $\left( \frac{1}{{{a}^{3}}-{{b}^{3}}}\cdot \frac{ac+ad-bc-bd}{1} \right)-\frac{c-d}{{{a}^{2}}+ab+{{b}^{2}}}=$ $\frac{2d}{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}$.
