\begin{align}
\prod_{n=3}^{\infty} \frac{(n^3+3n)^2}{n^6-64}
&= \prod_{n=3}^{\infty} \frac{n^2(n^2+3)^2}{n^6-64}
\\ &= \prod_{n=3}^{\infty} \frac{n^2(n^2+3)^2}{(n^2-4)(n^4+4n^2+16)}
\\ &= \left(\prod_{n=3}^{\infty} \frac{n^2}{n^2-4}\right)\left(\prod_{n=3}^{\infty} \frac{(n^2+3)^2}{n^4+4n^2+16}\right)
\end{align}
consider (to find its roots and thus factorise it)
\begin{align}
n^4+4n^2+16&=0
\\ \iff n^2&=\frac{-4±\sqrt{16-64}}{2}
\\ &= -2±\sqrt{-12}
\\ &= -2±2i\sqrt3
\\ &= 4e^{i\pi(±\frac 2 3 +2k)} \ \forall \ k\in\mathbb{Z}
\\ \iff n&=(4e^{i\pi(±\frac 2 3 +2k)})^{\frac 1 2 }
\\ &= 2e^{i\pi(±\frac 1 3 +k)}
\end{align}
using de moivre’s theorem. we have a quartic so there are 4 roots (2 pairs of complex conjugates).
we get that the roots are \displaystyle n=2e^{±\frac{i\pi}{3}},2e^{±\frac{5i\pi}{3}}
\begin{align}
\therefore n^4+4n^2+16
&=(n-2e^{\frac{i\pi}{3}})(n-2e^{-\frac{i\pi}{3}})(n-2e^{\frac{5i\pi}{3}})(n-2e^{-\frac{5i\pi}{3}})
\\ &= (n^2-2(e^{\frac{i\pi}{3}}+e^{-\frac{i\pi}{3}})n+4e^{\frac{i\pi}{3}}e^{-\frac{i\pi}{3}})(n^2-2(e^{\frac{5i\pi}{3}}+e^{-\frac{5i\pi}{3}})n+4e^{\frac{5i\pi}{3}}e^{-\frac{5i\pi}{3}})
\\ &= (n^2-4n\cos\frac{\pi}{3}+4)(n^2-4n\cos\frac{5\pi}{3}+4)
\\ &= (n^2-2n+4)(n^2+2n+4)
\end{align}
alternatively (just spotted this much simpler alternative, still complex method was a nice alternative i think, although would be more useful in other situations), just note that
\begin{align}
n^4+4n^2+16
&= n^4+8n^2+16-4n^2
\\ &= (n^2+4)^2-(2n)^2
\\ &= (n^2-2n+4)(n^2+2n+4)
\end{align}
now we have
\begin{align}
\left(\prod_{n=3}^{\infty} \frac{n^2}{n^2-4}\right)\left(\prod_{n=3}^{\infty} \frac{(n^2+3)^2}{n^4+4n^2+16}\right)
&= \left(\prod_{n=3}^{\infty} \frac{n}{n-2}\right)\left(\prod_{n=3}^{\infty} \frac{n}{n+2}\right)\left(\prod_{n=3}^{\infty} \frac{(n^2+3)^2}{(n^2-2n+4)(n^2+2n+4)}\right)
\\ &= \left(\prod_{n=3}^{\infty} \frac{n}{n-2}\right)\left(\prod_{n=3}^{\infty} \frac{n}{n+2}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{n^2-2n+4}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{n^2+2n+4}\right)
\\ &=\left(\prod_{n=3}^{\infty} \frac{n}{n-2}\right)\left(\prod_{n=3}^{\infty} \frac{n}{n+2}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{(n-1)^2+3}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{(n+1)^2+3}\right)
\\ &=\left(\lim_{N\to\infty}\prod_{n=3}^{N}n\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n-2}\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N}n\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n+2}\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N}n^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n-1)^2+3}\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N}n^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n+1)^2+3}\right)
\\ &= \left(\lim_{N\to\infty}\prod_{n=5}^{N+2}n-2\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n-2}\right)\left(\lim_{N\to\infty}\prod_{n=1}^{N-2}n+2\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n+2}\right)\left(\lim_{N\to\infty}\prod_{n=4}^{N+1}(n-1)^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n-1)^2+3}\right)\left(\lim_{N\to\infty}\prod_{n=2}^{N-1}(n+1)^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n+1)^2+3}\right)
\\ &= \left(\lim_{N\to\infty}\frac 1 2 N(N-1)\right)\left(\lim_{N\to\infty}\frac{3\times4}{(N+1)(N+2)}\right)\left(\lim_{N\to\infty}\frac{N^2+3}{7}\right)\left(\lim_{N\to\infty}\frac{12}{(N+1)^2+3}\right)
\\ &= \lim_{N\to\infty} \frac{N\times(N-1)\times3\times4\times(N^2+3)\times12}{2\times (N+1)\times(N+2)\times7\times((N+1)^2+3)}
\end{align}
as N\to\infty, N+a\to N where a\in\mathbb{R}. so we can simplify this to:
\begin{align}
\lim_{N\to\infty} \frac{N\times(N-1)\times3\times4\times(N^2+3)\times12}{2\times (N+1)\times(N+2)\times7\times((N+1)^2+3)}
&=\frac{\cancel{N}\times \cancel{N}\times3\times4\times\cancel{(N^2+3)}\times12}{2\times \cancel{N}\times \cancel{N}\times7\times\cancel{(N^2+3)}}
\\ &= \frac{72}7
\end{align}
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