#### Answer

The provided function is an odd function, has symmetry about the origin only.

#### Work Step by Step

Step I: To check even or odd.
In the provided equation, if $f\left( -x \right)=f\left( x \right)$ then it is an even function, and if $f\left( -x \right)=-f\left( x \right)$ , then it is odd.
$\begin{align}
& f\left( x \right)={{x}^{3}}+x \\
& f\left( -x \right)={{\left( -x \right)}^{3}}+\left( -x \right) \\
& =-{{x}^{3}}-x \\
& =-\left( {{x}^{3}}+x \right) \\
& =-f\left( x \right)
\end{align}$
It is an odd function.
Step II: To check symmetry about the y-axis:
In the provided equation put $x=-x$ , if the equation remains the same, then it has symmetry about the y- axis.
$\begin{align}
& y={{x}^{3}}+x \\
& y={{\left( -x \right)}^{3}}+\left( -x \right) \\
& y=-{{x}^{3}}-x \\
\end{align}$
It is not the same as the given equation, hence it is not symmetric about the y-axis.
Step III: To check symmetry about the x-axis:
In the provided equation put $y=-y$ , if the equation remains the same, then it has symmetry about the x-axis.
$\begin{align}
& y={{x}^{3}}+x \\
& \left( -y \right)={{x}^{3}}+x \\
& -y={{x}^{3}}+x
\end{align}$
It is not the same as the given equation, hence it is not symmetric about the x-axis.
Step IV: To check symmetry about the origin:
In the provided equation put $x=-x,\text{ and }y=-y$ , if the equation remains the same, then it has symmetry about the origin.
$\begin{align}
& y={{x}^{3}}+x \\
& \left( -y \right)={{\left( -x \right)}^{3}}+\left( -x \right) \\
& -y=-{{x}^{3}}-x \\
& y={{x}^{3}}+x
\end{align}$
It is the same as provided equation, hence it has symmetry about the origin.
Thus, the given function is an odd function, has symmetry about the origin only.