smc

Let $P(x)=A{x}^3+B{x}^2+Cx+D$. 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 $$P(2)+P(-2)=12k+2k=\boxed{14k}$$