This expression can be solved by using the Sophie Germain's Identity (see %%HREF%%3863%%). To avoid a very long expression after factorization, let's handle each terms separately. Note that $324=4\times 3^4$, we have \begin{align*} &10^4+324 = (10^2+ 2\times 3^2+2\times 10\times 3)(10^2+ 2\times 3^2-2\times 10\times 3)=178\times 58\\ &22^4+324= (22^2+2\times 3^2+2\times 22\times 3)(22^2+2\times 3^2-2\times 22\times 3)= 634\times 370\\ &34^4+324= (34^2+2\times 3^2+2\times 34\times 3)(34^2+2\times 3^2-2\times 34\times 3)= 1378\times 970\\ &46^4+324= (46^2+2\times 3^2+2\times 46\times 3)(46^2+2\times 3^2-2\times 46\times 3)= 2410\times 1858\\ &58^4+324= (58^2+2\times 3^2+2\times 58\times 3)(58^2+2\times 3^2-2\times 58\times 3)= 3730\times 3034\\ &4^4+324= (4^2+2\times 3^2+2\times 4\times 3)(4^2+2\times 3^2-2\times 4\times 3)= 58\times 10\\ &16^4+324= (16^2+2\times 3^2+2\times 16\times 3)(16^2+2\times 3^2-2\times 16\times 3)= 370\times 178\\ &28^4+324= (28^2+2\times 3^2+2\times 28\times 3)(28^2+2\times 3^2-2\times 28\times 3)= 970\times 634\\ &40^4+324= (40^2+2\times 3^2+2\times 40\times 3)(40^2+2\times 3^2-2\times 40\times 3)= 1858\times 1378\\ &52^4+324= (52^2+2\times 3^2+2\times 52\times 3)(52^2+2\times 3^2-2\times 52\times 3)= 3034\times 2410 \end{align*} Therefore \begin{align*} &\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}\\ &=\frac{178\times 58\times 634\times 370\times 1378\times 970\times 2410\times 1858\times 3730\times 3034}{58\times 10\times 370\times178\times 970\times 634\times 1858\times 1378\times 3034\times 2410}\\ &=\frac{3730}{10}\\ &=\boxed{373} \end{align*}