Answer
${{t}^{2}}-t+1$
Work Step by Step
$\begin{align}
& \frac{{{t}^{3}}+{{1}^{3}}}{t+1}=\frac{\left( t+1 \right)\left( {{t}^{2}}-t+1 \right)}{t+1} \\
& ={{t}^{2}}-t+1
\end{align}$
Thus,
$\frac{{{t}^{3}}+1}{t+1}={{t}^{2}}-t+1$
Therefore, the simplified form of the expression $\frac{{{t}^{3}}+1}{t+1}$ is ${{t}^{2}}-t+1$.
