Answer
$x\displaystyle \in\left\{-\frac{1}{3}\right\}$
Work Step by Step
Multiply the equation with 3 (to get rid of fractions)
$3x^{3}-2x^{2}+8x+3=0$
We try to find rational zeros of $f(x)=3x^{3}-2x^{2}+8x+3$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 3}{\pm 1,\pm 3}$
Testing with synthetic division, ... we eventually try $x+\displaystyle \frac{1}{3}$
$\left.\begin{array}{l}
-1/3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 & -2 & 8 &3 \\\hline
& -1 & 1 & -3 \\\hline
3& -3 &9 & |\ \ 0 \end{array}$
$f(x)=(x+\displaystyle \frac{1}{3})(3x^{2}-3x+9)=(x+\frac{1}{3})\cdot 3(x^{2}-x+3)$
$=(3x+1)(x^{2}-x+3)$
The discriminant, $b^{2}-4ac=1-4(3)=-11$
of the equation $x^{2}-x+3=0$
is negative, so the equation has no real solutions.
No further factoring,
$f(x)=(3x+1)(x^{2}-x+3)=0$
when $x\displaystyle \in\left\{-\frac{1}{3}\right\}$
