Let \(\displaystyle{x}{2}={a}{\left({1}+{x}\right)}+{b}{\left({2}+{x}^{{{2}}}\right)}+{c}{\left(−{x}\right)}{x}^{{{2}}}={a}{\left({1}+{x}\right)}+{b}{\left({2}+{x}^{{{2}}}\right)}+{c}{\left(−{x}\right)}.\) Then comparing the coefficients of power of x we get
\(\displaystyle{a}+{2}{b}={0}⋯{\left({1}\right)}\)

\(\displaystyle{a}−{c}={0}⋯{\left({2}\right)}\)

\(\displaystyle{2}{b}={1}⋯{\left({3}\right)}.\)

From (3) we have b=1/2. From (1) and (2) we have \(\displaystyle{c}=−{2}{b}=−{1}{c}=−{2}{b}=−{1}\). Therefore a=c=−1. Thus x^{2} can be written as a linear combination of \(\displaystyle{\left({1}+{x}\right)},{\left({2}+{x}^{{{2}}}\right)}{\left({1}+{x}\right)},{\left({2}+{x}^{{{2}}}\right)}\) and −x.

\(\displaystyle{a}−{c}={0}⋯{\left({2}\right)}\)

\(\displaystyle{2}{b}={1}⋯{\left({3}\right)}.\)

From (3) we have b=1/2. From (1) and (2) we have \(\displaystyle{c}=−{2}{b}=−{1}{c}=−{2}{b}=−{1}\). Therefore a=c=−1. Thus x^{2} can be written as a linear combination of \(\displaystyle{\left({1}+{x}\right)},{\left({2}+{x}^{{{2}}}\right)}{\left({1}+{x}\right)},{\left({2}+{x}^{{{2}}}\right)}\) and −x.