roots of: 1/3x^3-1/2x^2+1/3x+1/3=0

1/3x^3-1/2x^2+1/3x+1/3=0    let's multiply through by the common denominator of 2 and 3  = 6.....so we have

2x^3 - 3x^2 + 2x + 2 = 0

This will not factor......I can see that one solution is between  -1 and 0....let's use the on-site solver to find all of them.....

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{1}}-{\mathtt{1}}{i}\\
{\mathtt{x}} = {i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{1}}{\mathtt{\,-\,}}{i}\\
{\mathtt{x}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}\\
{\mathtt{x}} = -{\mathtt{0.5}}\\
\end{array} \right\}$$

I actually should have used the Rational Zeroes Theorem to find the "real" solution.....the other two could have been found by using the Quadratic formula.