Answer
The function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function and is symmetric about the origin.
Work Step by Step
In order to check if the function is even, odd, or neither, substitute $x$ by – $x$ and evaluate the value of $f\left( -x \right)$.
$\begin{align}
& f\left( -x \right)={{\left( -x \right)}^{3}}-5\left( -x \right) \\
& =\left( -x \right)\left( -x \right)\left( -x \right)-5\left( -x \right) \\
& =-{{x}^{3}}+5x \\
& =-\left( {{x}^{3}}-5x \right)
\end{align}$
Since, $f\left( x \right)={{x}^{3}}-5\left( x \right)$, therefore $f\left( -x \right)=-f\left( x \right)$.
Now as $f\left( -x \right)=-f\left( x \right)$, it gives that the function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function
The function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function, therefore by definition of an odd function, $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is symmetric about the origin.
Hence, the function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function and is symmetric about the origin.
