Answer
See below
Work Step by Step
Consider $p_0(x)=x^3\\
p_1(x)=1+x^3\\
p_2(x)=x+x^3\\
p_3(x)=x^2+x^3
\in P_3$
Obtain the matrix $A=\begin{bmatrix}
x^3 & 1+x^3 & x+x^3 & x^2+x^3\\ 3x^2 & 3x^2 & 1+3x^2 & 2x+3x^2 \\6x & 6x & 6x & 2+6x\\6 & 6 & 6 & 6
\end{bmatrix}\approx \begin{bmatrix}
-1 & 1+x^3 & x+x^3 & x^2+x^3\\ 0 & 3x^2 & 1+3x^2 & 2x+3x^2 \\0 & 6x & 6x & 2+6x\\0 & 6 & 6 & 6
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1-x & x+x^3 & x^2+x^3\\ 0 & -1 & 1+3x^2 & 2x+3x^2 \\0 &0 & 6x & 2+6x\\0 & 0 & 6 & 6
\end{bmatrix}\approx \begin{bmatrix}
-1 & 1-x & x-x^2 & x^2+x^3\\ 0 & -1 & 1-2x & 2x+3x^2 \\0 &0 & -2 & 2+6x\\0 & 0 &0 & 6
\end{bmatrix} \\
\rightarrow \det =-12$
Since $\det (A(x))=W[p_0,p_1,p_2,p_3]\\
\rightarrow W[p_0,p_1,p_2,p_3]=-12\ne 0$
Thus, the set of vectors $\{p_0,p_1,p_2,p_3\}$ is linearly independent on any interval. Then the set is also a linearly independent set in $P_3$
We know that $dim [P_3]=4 \rightarrow \{p_0,p_1,p_2,p_3\}$ is a basic for $P_3$
Hence, $\{p_0,p_1,p_2,p_3\}$ is a basic for $P_3(R)$ whose elements all have the same degree.
