Answer
$ \dfrac{4x-7}{x^{2} -x+1}$
Work Step by Step
Factor $x^3+1$ using the sum of two cubes formula:
$\displaystyle \begin{array}{ l l }
=\dfrac{4}{x+1} +\dfrac{1}{x^{2} -x+1} -\dfrac{12}{( x+1)\left( x^{2} -x+1\right)} & \begin{array}{{>{\displaystyle}l}}
\end{array}\\
\end{array}\\$
Make the expressions similar using their LCD which is $(x+1)(x^2-x+1)$.
$=\dfrac{4}{x+1} \cdot \dfrac{x^{2} -x+1}{x^{2} -x+1} +\dfrac{1}{x^{2} -x+1} \cdot \dfrac{x+1}{x+1} -\dfrac{12}{( x+1)\left( x^{2} -x+1\right)}$
$\\
=\dfrac{4\left( x^{2} -x+1\right)}{(x+1)\left( x^{2} -x+1\right)} +\dfrac{1( x+1)}{(x^{2} -x+1)( x+1)} -\dfrac{12}{( x+1)\left( x^{2} -x+1\right)}$
Apply the Distributive Property;
$=\dfrac{4x^{2} -4x+4}{( x+1)\left( x^{2} -x+1\right)} +\dfrac{x+1}{( x+1)\left( x^{2} -x+1\right)} -\dfrac{12}{( x+1)\left( x^{2} -x+1\right)}$
Combine like terms
$=\dfrac{4x^{2} -4x+4\ +\ ( x+1) \ -\ 12}{( x+1)\left( x^{2} -x+1\right)}\\$
$=\dfrac{4x^{2} +( -4x+x) +( 4\ +\ 1-12)}{( x+1)\left( x^{2} -x+1\right)} $
$=\dfrac{4x^{2} -3x-7}{( x+1)\left( x^{2} -x+1\right)} $
Factor the numerator:
$=\dfrac{( 4x-7)( x+1)}{( x+1)\left( x^{2} -x+1\right)} $
Cancel the common factors:
$\require{cancel}
=\dfrac{( 4x-7)\cancel{( x+1)}}{\cancel{( x+1)}\left( x^{2} -x+1\right)} $
$=\dfrac{4x-7}{x^2-x+1}$
