1. Let the roots of $x^4-3x^3-2x^2+3x+1=0$ be $\alpha,\beta,\gamma,\delta$.

We need to find: $\alpha^3+\beta^3+\gamma^3+\delta^3.$

2. Use Newton's sums / symmetric sums.

For the monic polynomial $x^4-3x^3-2x^2+3x+1=0,$ by Vieta's formulas: $\alpha+\beta+\gamma+\delta=3$ $\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=-2$ $\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-3$ $\alpha\beta\gamma\delta=1.$

Let $p_1=\alpha+\beta+\gamma+\delta,$ $p_2=\alpha^2+\beta^2+\gamma^2+\delta^2,$ $p_3=\alpha^3+\beta^3+\gamma^3+\delta^3.$

We already have $p_1=3.$

3. First compute $p_2$ using $p_2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta).$ So, $p_2=3^2-2(-2)=9+4=13.$

4. Now use Newton's sum for cubes: $p_3-e_1p_2+e_2p_1-3e_3=0,$ where $e_1=3,\quad e_2=-2,\quad e_3=-3.$ Thus, $p_3-3(13)+(-2)(3)-3(-3)=0.$ Simplify: $p_3-39-6+9=0$ $p_3-36=0$ $p_3=36.$

5. Therefore, the sum of the cubes of all the roots is $\boxed{36}.$