2013-II-31

Ans: B
$\begin{array}{rcl}
a^2+4a+4 & = & (a+2)^2 \\
a^2-4 & = & (a+2)(a-2) \\
a^3 + 8 & = & (a+2)(a^2-2a+4) \\ \hline
\therefore \text{L.C.M.} & = & (a-2)(a+2)^2(a^2-2a+4)
\end{array}$