Answer
$S$ is not a basis for $P_3$.
Work Step by Step
The set $S=\left\{4-t, t^{3}, 6 t^{2}, t^{3}+3 t, 4 t-1\right\}$ is not a basis for $P_3$. Indeed, since $P_3$ has dimension $4$, then any basis for $P_3$ must contain $4$ vectors. But, here $S$ contains $ 5$ vectors and hence $S$ is not a basis for $P_3$.
