jmc

If we multiply $(x^2-k)$ by $(x+k)$, we get $x^3+k{x}^2-kx-k^2$. We can now factor out a $k$ from the last three terms of this expression, which gives us $x^3+k(x^2-x-k)$. When we set this equal to the right side of the original equation $x^3+k(x^2-x-5)$, we get $x^3+k(x^2-x-k)=x^3+k(x^2-x-5)$. A careful comparison of the two sides of this equation reveals that $k$ must be 5 (consider the constant terms). Alternatively, we could multiply out both sides of the equation and get $x^3+k{x}^2-kx-k^2=x^3+k{x}^2-kx-5k$. The left side and the right side are the exact same when $k^2=5k$, so $k=\boxed{5}$.