Answer
$S$ spans $P_3$.
Work Step by Step
Consider the combination
$$a(x^2-2x)+b(x^3+8)+c(x^3-x^2)+d(x^2-4)=0.$$
Comparing the coefficients we get the system
\begin{align*}
b+c&=0\\
a+b-c+d&=0\\
-2a&=0\\
8b-4d&=0.
\end{align*}
The coefficient matrix
$$\left[ \begin{array} {cc}0&1&1&0\\1&1&1&1\\-2&0&0&0\\
0&8&0&-4 \end{array} \right]$$
has zero determinant since
$$ \left| \begin{array} {cc}0&1&1&0\\1&1&1&1\\-2&0&0&0\\
0&8&0&-4 \end{array} \right|=-2\left| \begin{array} {cc} 1&1&0\\ 1&1&1\\
8&0&-4 \end{array} \right|=-2(-4-(-4))=0.$$
Hence, $S$ is linearly independent set of vectros and since $P_3$ has dimension $4$ then $S$ spans $P_3$.
