Use partial fractions to find the inverse Laplace transform of $F(s) =\large \frac{s+1}{s^3+5s^2+6s}$. Then use $u = s+1$ to show that $F(s) = G(u) = \large\frac{u}{(u-1)(u+1)(u+2)}$.
Use partial fractions to find the inverse Laplace transform of $G(s)$, and then use the translation theorem to show that the inverse transforms of $F(s)$ and $G(s+1)$ are the same.
I found the inverse Laplace transforms of $F(s)$ and $G(s)$ using partial fractions as:
$\mathcal{L}^{-1}\{F(s)\} = f(t) = \frac{1}{6}-\frac{2}{3}e^{-3t}+\frac{1}{2}e^{-2t}$
$\mathcal{L}^{-1}\{G(s)\} = g(t) = \frac{1}{6}e^{t}+\frac{1}{2}e^{-t}-\frac{2}{3}e^{-2t}$
Now I'm having trouble using the translation theorem to show that $F(s)$ and $G(s+1)$ are the same. Can some explain exactly what the translation theorem allows us to conclude and how it can possibly relate to this problem?