Answer
$S$ is a basis for $P_3$.
Work Step by Step
Let $S$ be given by $$
S=\left\{1-t, 2 t+3 t^{2}, t^{2}-2 t^{3}, 2+t^{3}\right\}.
$$
Consider the combination
$$a(1-t)+b(2 t+3 t^{2})+c(t^{2}-2 t^{3})+d(2+t^{3})=0, \quad a,b,c,d\in R.$$
Which yields the following system of equations
\begin{align*}
a+2d&=0\\
-a+2b&=0\\
3b+c&=0\\
-2c+d&=0.
\end{align*}
The coefficient matrix
$$\left[ \begin {array}{cccc} 1&0&0&2\\ -1&2&070\\0&3&1&0\\0&0&-2&1\end {array}
\right] $$
has non zero determinant and hence there exist a unique solution for the above system; that is, the trivial solution.
Then, $S$ is linearly independent set of vectors. Since, $P_3$ has dimension $4$ then, by Theorem 4.12 $S$ is a basis for $P_3$.
