Answer
$\text{The factored form of the given expression is}:$
$\left[( x+1)\left( x^{2} -x+1\right)\right]\left[( x-1)\left( x^{2} +x+1\right)\right]$
Work Step by Step
$\displaystyle \begin{array}{|l|l|}
\hline
=x^{6} -1^{6} & \begin{array}{{>{\displaystyle}l}}
\mathrm{We\ can\ re-write} \ 1\ \mathrm{as}\\
1^{6} \ \mathrm{since} \ 1^{a} =1\\
\mathrm{a\ is\ any\ integer\ greater\ than\ 0}
\end{array}\\
& \\
\hline
=\left( x^{3}\right)^{2} -\left( 1^{3}\right)^{2} & \mathrm{Apply\ the\ rule} \ a^{n\cdot m} =\left( a^{n}\right)^{m}\\
& \\
\hline
=\left( x^{3} +1\right)\left( x^{3} -1\right) & \begin{array}{{>{\displaystyle}l}}
\mathrm{We\ can\ apply\ the\ difference\ of}\\
\mathrm{squares\ here} :\ \\
a^{2} -b^{2} =( a+b)( a-b)\\
\mathrm{In\ this\ case} ,\ a\ =\ x^{3} \ \mathrm{and} \ \\
b\ =\ 1
\end{array}\\
& \\
\hline
=\left[( x+1)\left( x^{2} -x+1\right)\right]\left( x^{3} -1\right) & \begin{array}{{>{\displaystyle}l}}
\mathrm{Apply\ the\ sum\ of\ cubes\ here} :\\
a^{3} +b^{3} =( a+b)\left( a^{2} -ab+b^{3}\right)
\end{array}\\
=\left[( x+1)\left( x^{2} -x+1\right)\right]\left[( x-1)\left( x^{2} +x+1\right)\right] & \begin{array}{{>{\displaystyle}l}}
\mathrm{Apply\ the\ difference\ of\ cubes\ here}\\
a^{3} -b^{3} =( a-b)\left( a^{2} +ab+b^{2}\right)
\end{array}\\
\hline
\end{array}$
