Answer
The distance, d, from $\left( x,y \right)$ to the origin is $d=\sqrt{{{x}^{2}}+{{y}^{2}}}$. If $\left( x,y \right)$ lies on the graph of $y={{x}^{3}}$ , we can replace y with ${{x}^{3}}$ in the formula for d. Thus $d\left( x \right)=\sqrt{{{x}^{2}}+{{x}^{6}}}$.
Work Step by Step
Consider the statement: $\left( x,y \right)$ is assumed to be a point on the graph of $y={{x}^{3}}$.
Distance, d of a point $\left( x,y \right)$ from the origin $\left( 0,0 \right)$
Use distance formula: $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ ,
Substitute the coordinates of point $P\left( x,y \right)$ and the origin $\left( 0,0 \right)$ in $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$\begin{align}
& d=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{y}^{2}}}
\end{align}$
Substituting the value of y from $y={{x}^{3}}$ , d becomes:
$\begin{align}
& d\left( x \right)=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{\left( {{x}^{3}} \right)}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{x}^{6}}}
\end{align}$
Hence, the distance, d, from $\left( x,y \right)$ to the origin is $d=\sqrt{{{x}^{2}}+{{y}^{2}}}$. If $\left( x,y \right)$ lies on the graph of $y={{x}^{3}}$ , we can replace y with ${{x}^{3}}$ in the formula for d. Thus $d\left( x \right)=\sqrt{{{x}^{2}}+{{x}^{6}}}$.
