Use the Distance Formula:
\(\displaystyle{d}=\sqrt{{{\left({x}{2}-{x}{1}\right)}^{{2}}+{\left({y}{2}-{y}{1}\right)}^{{2}}}}\)

The distance between (1,0) and (x,y) is: \(\displaystyle{d}{1}=\sqrt{{{\left({x}-{1}\right)}^{{2}}+{\left({y}-{0}\right)}^{{2}}}}\)

\(\displaystyle{d}{1}=\sqrt{{{\left({x}-{1}\right)}^{{2}}+{y}^{{2}}}}\)

Since \(\displaystyle{y}={x}^{{3}}\), we substitute:

\(\displaystyle{d}{1}=\sqrt{{{\left({x}^{{2}}-{2}{x}+{1}\right)}+{\left({x}^{{3}}\right)}^{{2}}}}\)

\(\displaystyle{d}{1}=\sqrt{{{\left({x}^{{2}}-{2}{x}+{1}\right)}+{x}^{{6}}}}\)

\(\displaystyle{d}{1}=\sqrt{{{\left({x}^{{6}}\right)}+{x}^{{2}}-{2}{x}+{1}}}\)

Similarly, the distance between (0,1) and (x,y) is: \(\displaystyle{d}{2}=\sqrt{{{\left({x}-{0}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}}}\)

\(\displaystyle{d}{2}=\sqrt{{{\left({x}^{{2}}\right)}+{\left({y}-{1}\right)}^{{2}}}}\)

Since \(\displaystyle{y}={x}^{{3}}\), we substitute: \(\displaystyle{d}{2}=\sqrt{{{\left({x}^{{2}}+{\left({x}^{{3}}-{1}\right)}^{{2}}\right)}}}\)

\(\displaystyle{d}{2}=\sqrt{{{x}^{{2}}+{\left({x}^{{6}}-{2}{x}^{{3}}+{1}\right)}}}\)

\(\displaystyle{d}{2}=\sqrt{{{x}^{{6}}-{2}{x}^{{3}}+{x}^{{2}}+{1}}}\)

The distance between (1,0) and (x,y) is: \(\displaystyle{d}{1}=\sqrt{{{\left({x}-{1}\right)}^{{2}}+{\left({y}-{0}\right)}^{{2}}}}\)

\(\displaystyle{d}{1}=\sqrt{{{\left({x}-{1}\right)}^{{2}}+{y}^{{2}}}}\)

Since \(\displaystyle{y}={x}^{{3}}\), we substitute:

\(\displaystyle{d}{1}=\sqrt{{{\left({x}^{{2}}-{2}{x}+{1}\right)}+{\left({x}^{{3}}\right)}^{{2}}}}\)

\(\displaystyle{d}{1}=\sqrt{{{\left({x}^{{2}}-{2}{x}+{1}\right)}+{x}^{{6}}}}\)

\(\displaystyle{d}{1}=\sqrt{{{\left({x}^{{6}}\right)}+{x}^{{2}}-{2}{x}+{1}}}\)

Similarly, the distance between (0,1) and (x,y) is: \(\displaystyle{d}{2}=\sqrt{{{\left({x}-{0}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}}}\)

\(\displaystyle{d}{2}=\sqrt{{{\left({x}^{{2}}\right)}+{\left({y}-{1}\right)}^{{2}}}}\)

Since \(\displaystyle{y}={x}^{{3}}\), we substitute: \(\displaystyle{d}{2}=\sqrt{{{\left({x}^{{2}}+{\left({x}^{{3}}-{1}\right)}^{{2}}\right)}}}\)

\(\displaystyle{d}{2}=\sqrt{{{x}^{{2}}+{\left({x}^{{6}}-{2}{x}^{{3}}+{1}\right)}}}\)

\(\displaystyle{d}{2}=\sqrt{{{x}^{{6}}-{2}{x}^{{3}}+{x}^{{2}}+{1}}}\)