For any \(\displaystyle{a}\in{R}\), we have,

\(\displaystyle{\left({a}+{a}\right)}={\left({a}+{a}\right)}^{{2}}\)

\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={\left({2}{a}\right)}^{{2}}\)

\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={4}{a}^{{2}}\)

\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={a}^{{2}}+{a}^{{2}}+{a}^{{2}}+{a}^{{2}}\)

\(\displaystyle\Rightarrow{\left({a}+{a}\right)}={a}+{a}+{a}+{a}\ldots{\left({1}\right)}\)

Now, let \(\displaystyle{a},{b}\in{R}\)

\(\displaystyle{a}+{b}={\left({a}+{b}\right)}^{{2}}\)

\(\displaystyle={a}^{{2}}+{a}{b}+{b}{a}+{b}^{{2}}\)

\(\displaystyle={a}+{a}{b}+{b}{a}+{b}\)

\(\displaystyle\Rightarrow{a}{b}+{b}{a}={0}\)

Adding ba on both sides of above equation, we get,

\(ab+(ba+ba)=ba\)

From equation 1, we can observe that same terms on both sides of equation get cancelled out.

So, applying same observation on equation 2, we get, \(ab=ba\)

Hence, Ring R is commutative.