$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$
$\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$
First, note that we can pull the exponents out of every factor, since they are all squared. This results in
$\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$
Now, multiplying the numerators together gives
$\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}$,
which simplifies to $\boxed{1\textbf{ (C)}}$.