Answer
$\frac{1}{3},i\sqrt{\frac{8}{3}},-i\sqrt{\frac{8}{3}}$
Work Step by Step
$f\left( x \right)=\left( x-\frac{1}{3} \right)\left( {{x}^{2}}+6 \right)$
And,
$\text{g}\left( x \right)=\left( x-\frac{1}{3} \right)\left( {{x}^{2}}-\frac{2}{3} \right)$
At $x=a$, the value of the function $f\left( x \right)$ will be,
$\begin{align}
& f\left( x \right)=\left( x-\frac{1}{3} \right)\left( {{a}^{2}}+6 \right) \\
& f\left( a \right)=\left( a-\frac{1}{3} \right)\left( {{x}^{2}}+6 \right) \\
\end{align}$
At $x=a$, the value of the function $g\left( x \right)$ will be,
$\begin{align}
& \text{g}\left( x \right)=\left( x-\frac{1}{3} \right)\left( {{x}^{2}}-\frac{2}{3} \right) \\
& \text{g}\left( a \right)=\left( a-\frac{1}{3} \right)\left( {{a}^{2}}-\frac{2}{3} \right) \\
\end{align}$
Using:
$\left( f+g \right)\left( a \right)=f\left( a \right)+g\left( a \right)$
Put the values of $f\left( a \right)\text{ and }g\left( a \right)$ in the above equation,
$\begin{align}
& \left( f+g \right)\left( a \right)=f\left( a \right)+g\left( a \right) \\
& \left( f+g \right)\left( a \right)=\left( a-\frac{1}{3} \right)\left( {{a}^{2}}+6 \right)+\left( a-\frac{1}{3} \right)\left( {{a}^{2}}-\frac{2}{3} \right) \\
\end{align}$
And from the equation, $\left( f+g \right)\left( a \right)=0$, so
$\begin{align}
& \left( f+g \right)\left( a \right)=\left( a-\frac{1}{3} \right)\left( {{a}^{2}}+6 \right)+\left( a-\frac{1}{3} \right)\left( {{a}^{2}}-\frac{2}{3} \right) \\
& 0=\left( a-\frac{1}{3} \right)\left( {{a}^{2}}+6 \right)+\left( a-\frac{1}{3} \right)\left( {{a}^{2}}-\frac{2}{3} \right)
\end{align}$
$\begin{align}
& \left( a-\frac{1}{3} \right)\left( {{a}^{2}}+6 \right)+\left( a-\frac{1}{3} \right)\left( {{a}^{2}}-\frac{2}{3} \right)=0 \\
& \left( a-\frac{1}{3} \right)\left( {{a}^{2}}+6+{{a}^{2}}-\frac{2}{3} \right)=0 \\
& \left( a-\frac{1}{3} \right)\left( 2{{a}^{2}}+\frac{16}{3} \right)=0
\end{align}$
For $\left( a-\frac{1}{3} \right)=0$,
$\begin{align}
& a-\frac{1}{3}=0 \\
& a=\frac{1}{3} \\
\end{align}$
For $\left( 2{{a}^{2}}+\frac{16}{3} \right)=0$,
$\begin{align}
& 2{{a}^{2}}+\frac{16}{3}=0 \\
& 2{{a}^{2}}=-\frac{16}{3} \\
& {{a}^{2}}=-\frac{16}{6}
\end{align}$
This gives:
$\begin{align}
& {{a}^{2}}=-\frac{8}{3} \\
& a=\pm i\sqrt{\frac{8}{3}} \\
\end{align}$
Thus, the answer is: $\frac{1}{3},i\sqrt{\frac{8}{3}},-i\sqrt{\frac{8}{3}}$.
