Answer
$\displaystyle \begin{array}{ c l }
\mathrm{Step\ one} & \mathrm{Equivalent\ Expressions\ Property}\\
& \\
\mathrm{Step\ two} & \mathrm{Associative\ property}\\
& \\
\mathrm{Step\ three} & \mathrm{Binomial\ sum\ product\ property}\\
& \\
\mathrm{Step\ four} & \mathrm{Difference\ of\ Squares}\\
& \\
\mathrm{Step\ five} & \mathrm{Commutative\ Property}
\end{array}$
Work Step by Step
$\displaystyle \begin{array}{ c l }
\mathrm{Step\ one} & x^{2} \ \mathrm{is\ expanded\ to} \ 2x^{2} -x^{2}\\
& \\
\mathrm{Step\ two} & \begin{array}{{>{\displaystyle}l}}
\mathrm{Associative\ property\ is\ applied\ here}\\
a+b+c\ =( a+b) +c=a+( b+c)
\end{array}\\
& \\
\mathrm{Step\ three} & \begin{array}{{>{\displaystyle}l}}
\mathrm{Special\ product\ property\ is\ applied\ here}\\
( a+b)^{2} =\left( a^{2} +2ab+b^{2}\right)\\
\mathrm{In\ this\ case} \ a\ =\ x^{2} ,\ b\ =\ 1
\end{array}\\
& \\
\mathrm{Step\ four} & \begin{array}{{>{\displaystyle}l}}
\mathrm{Difference\ of\ Squares\ is\ applied\ here}\\
a^{2} +b^{2} =( a\ +\ b)( a\ -\ b)
\end{array}\\
& \\
\mathrm{Step\ five} & \begin{array}{{>{\displaystyle}l}}
\mathrm{Commutative\ Property\ States} :\\
a+b\ =\ b+a
\end{array}
\end{array}$
