$f(t)=1+t-\dfrac{8}{3}\displaystyle\int_{0}^{t}(\tau-t)^3f(\tau) \ \mathrm d\tau \quad , f(t)=?$ $$
f(t)=1+t-\dfrac{8}{3}\displaystyle\int_{0}^{t}(\tau-
t)^3f(\tau)  \ \mathrm d\tau
$$
According to the convolution theorem, $\displaystyle\int_{0}^{t}(\tau-
t)^3f(\tau)d\tau$ = $f(t) * t^3$ (I think)and its transform is equal to $F(s)=\frac{3!}{s^4}$
Finding the transform of the entire equation gives me 
$$
F(s)=\frac{1}{s}+\frac{1}{s^2}-F(s)\frac{16}{s^4}
$$
I got $F(s) = \dfrac{s^5+s^4}{s^2(s^4+16)}$, but I'm not sure if it is right. And even if it is, I don't know how to find the inverse transform. 
I was wrong $g(t-\tau) = g(t)$
In the problem $(\tau-t)^3 = -(t-\tau)^3 = -t^3$
In the end the denominator is $s^4 - 16$ which is much easier to deal with. Sorry for my oversight. 
 A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
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$\ds{\fermi\pars{t}
     =
     1 + t - {8 \over 3}\int_{0}^{t}\pars{\tau-t}^{3}\fermi\pars{\tau}\,\dd\tau\,,
\qquad\tilde{\rm f}\pars{s} = \int_{0}^{\infty}\expo{-st}\fermi\pars{t}\,\dd t}$

\begin{align}
\tilde{\rm f}\pars{s}&=
{1 \over s} + {1 \over s^{2}} - {8 \over 3}\int_{0}^{\infty}\dd t\,\expo{-st}
\int_{0}^{t}\pars{\tau-t}^{3}\fermi\pars{\tau}\,\dd\tau
\\[3mm]&=
{1 \over s} + {1 \over s^{2}}
+ {8 \over 3}\int_{0}^{\infty}\dd\tau\,\fermi\pars{\tau}
\int_{\tau}^{\infty}\pars{t - \tau}^{3}\expo{-st}\,\dd t\tag{1}
\\[3mm]&=
{1 \over s} + {1 \over s^{2}}
+ {8 \over 3}\int_{0}^{\infty}\dd\tau\,\fermi\pars{\tau}
\int_{0}^{\infty}t^{3}\expo{-s\pars{t + \tau}}\,\dd t
\\[3mm]&=
{1 \over s} + {1 \over s^{2}}
+ {8 \over 3}\,
\overbrace{\int_{0}^{\infty}\expo{-s\tau}\fermi\pars{\tau}\dd\tau}^{\ds{=\ \tilde{\rm f}\pars{s}}}\
\overbrace{\int_{0}^{\infty}t^{3}\expo{-st}\,\dd t}^{\ds{=\ 3!/s^{4}}}
\end{align}

$$
\tilde{\rm f}\pars{s}
=
{1 \over s} + {1 \over s^{2}} + {16 \over s^{4}}\,\tilde{\rm f}\pars{s}
\qquad\imp\qquad
\tilde{\rm f}\pars{s} = {s^{3} + s^{2} \over s^{4} - 16}
$$
The roots of $s^{4} - 16 = 0$ are given by $s_{n} = 2\expo{\ic n\pi/2}$ with
$n = 0, 1, 2, 3$:
$$
s_{0} = 2\,,\quad s_{1} = 2\ic\,,\quad s_{2} = -2\,,\quad s_{3} = -2\ic
\qquad\mbox{where}\quad \pars{s_{n}^{3} + s_{n}^{2}} \not= 0\,,\quad n = 0, 1, 2, 3
$$

With $\gamma > 2$:
\begin{align}
\fermi\pars{t}&=
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}\tilde{\rm f}\pars{s}\expo{st}\,
{\dd s \over 2\pi\ic}
=
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}{\dd s \over 2\pi\ic}\,
{\pars{s^{3} + s^{2}}\expo{st} \over s^{4} - 16}
=
\sum_{n = 0}^{3}\lim_{s \to s_{n}}{\pars{s - s_{n}}\pars{s^{3} + s^{2}}\expo{st} \over s^{4} - 16}
\\[3mm]&=
\sum_{n = 0}^{3}{\pars{s_{n}^{3} + s_{n}^{2}}\expo{s_{n}t} \over 4s_{n}^{3}}
=
{1 \over 4}\sum_{n = 0}^{3}\pars{1 + {1 \over s_{n}}}\expo{s_{n}t}
=
{1 \over 4}\sum_{n = 0}^{3}\expo{s_{n}t}
+ {1 \over 4}\sum_{n = 0}^{3}{\expo{s_{n}t} \over s_{n}}
\\[3mm]&=
{1 \over 4}\pars{\expo{2t} + \expo{2\ic t} + \expo{-2t} + \expo{-2\ic t}}
+
{1 \over 4}\pars{{\expo{2t} \over 2} + {\expo{2\ic t} \over 2\ic}
+ {\expo{-2t} \over -2} + {\expo{-2\ic t} \over -2\ic}}
\\[3mm]&=
{1 \over 2}\bracks{\cos\pars{2t} + \cosh\pars{2t}}
+
{1 \over 4}\bracks{\sin\pars{2t} + \sinh\pars{2t}}
\end{align}

$$\color{#0000ff}{\large%
\fermi\pars{t}
\color{#000000}{\ =\ }
{1 \over 2}\bracks{\cos\pars{2t} + \cosh\pars{2t}}
+
{1 \over 4}\bracks{\sin\pars{2t} + \sinh\pars{2t}}}
$$
