91Ó°ÊÓ

Multiply both sides of the equation by $3$.

$\begin{array}{rcl}3(x^3-\frac{2}{3}{x}^2+\frac{8}{3}x+1)& =& 3·0\\ 3x^3-2x^2+8x+3& =& 0\end{array}$

Now the corresponding function is $f\left(x\right)=3x^3-2x^2+8x+3$.

Now all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $3$are

$p:Â\pm 1,Â\pm 3$

The factors of the leading coefficient $3$are

$q:Â\pm 1,Â\pm 3$

So the possible rational zeros are

$\frac{p}{q}:Â\pm 1,Â\pm 3,Â\pm \frac{1}{3}$

For the function $f\left(x\right)=3x^3-2x^2+8x+3$, the function value $f(-3)$is given as

$$\begin{gathered} f(-3)=3{(-3)}^3-2{(-3)}^2+8(-3)+3 \\ f(-3)=-81-18-24+3 \\ f(-3)=-120 \end{gathered}$$

The function value $f(-1)$is given as

role="math" $$\begin{gathered} f(-1)=3{(-1)}^3-2{(-1)}^2+8(-1)+3 \\ f(-1)=-3-2-8+3 \\ f(-1)=-10 \end{gathered}$$

The function value $f(-\frac{1}{3})$is given as

$$\begin{gathered} f(-\frac{1}{3})=3{(-\frac{1}{3})}^3-2{(-\frac{1}{3})}^2+8(-\frac{1}{3})+3 \\ f(-\frac{1}{3})=-\frac{1}{9}-\frac{2}{9}-\frac{8}{3}+3 \\ f(-\frac{1}{3})=\frac{-1-2-24}{9}+3 \\ f(-\frac{1}{3})=-3+3 \\ f(-\frac{1}{3})=0 \end{gathered}$$

As $f(-\frac{1}{3})=0$, so $x=-\frac{1}{3}$is a zero.

So on performing synthetic division we get

So the depressed polynomial is $x^2-x+3$. The polynomial is prime and it does not have any real solution.

So the only real solution is $x=-\frac{1}{3}$.