Let $P$ be a cubic polynomial with $P(0) = k$, $P(1) = 2k$, and $P(-1) = 3k$.  What is $P(2) + P(-2)$ ?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k$
Let $P(x) = Ax^3+Bx^2 + Cx+D$.   Plugging in $0$ for $x$, we find $D=k$, and plugging in $1$ and $-1$ for $x$, we obtain the following equations:
\[A+B+C+k=2k\]
\[-A+B-C+k=3k\]
Adding these two equations together, we get
\[2B=3k\]
If we plug in $2$ and $-2$ in for $x$, we find that 
\[P(2)+P(-2) = 8A+4B+2C+k+(-8A+4B-2C+k)=8B+2k\]
Multiplying the third equation by $4$ and adding $2k$ gives us our desired result, so 
\[P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}\]