Observe that
\begin{align}
(2x)^{3y}-(2x)^{y} \\= (2x)^{y+2y}-(2x)^{y}\\=(2x)^{y}[(2x)^{2y}-1]
\end{align}
0
Bumbble Comm
On
Option C is correct obviously because we can see that
\begin{align}
(2x)^{3y}-(2x)^{y} \\= (2x)^{y+2y}-(2x)^{y}\\=(2x)^{y}[(2x)^{2y}-1]
\end{align}
Option B is wrong because if we take out $\ 2^y $ common then we will get
$\ 2^y{[2^{2y} . x^{3y}-x^y}]$
No doubt Option $\,\,\,\,C\,\,\,\,\,$
Note that $a^{m+n}=a^m \cdot a^n$
Observe that \begin{align} (2x)^{3y}-(2x)^{y} \\= (2x)^{y+2y}-(2x)^{y}\\=(2x)^{y}[(2x)^{2y}-1] \end{align}