Answer
See answer below
Work Step by Step
Assume that
$p_0(x)=1$
$p_1(x)=x$
$p_2(x)=x^2$
$p_3(x)=x^3$
Let's take: $W[p_0,p_1,p_2,p_3](x)=\begin{vmatrix}
1 & x &x^2 & x^3\\
0 & 1 & 2x & 3x^2 \\
0 & 0 & 2 & 6x \\
0 & 0 & 0 & 6
\end{vmatrix}=12$
Since $W[p_0,p_1,p_2,p_3](x)\ne0$ the set of vectors is linearly independent on any interval.
We can see any $p_x=a_0+a_1x+a_2x^2+a_3x^3$ in $P^3$ is a linear combination of $p_0,p_1,p_2,p_3$ which means $\{ p_0,p_1, p_2, p_3\}$ is a spanning set for $P_3$
Thus, $ \{ p_0,p_1,p_2,p_3 \}$ is a basis for $P_3$ and $dimP_3 =4$
