$\frac{(10^4+324)(22^4+324)\cdots(58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$ 
From AIME 1987, compute $$\frac{(10^4+324)(22^4+324)\cdots (58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$$

So basically the way used to solve this is by Sophie Germain's Identity which is $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$
My question is , how is possible a student to solve this question without knowing this identity? Is there another way to solve this?

(Without using Sophie Germain's identity, per OP's request.)
The given expression is $\,f(7, 5)\,$ where $\,\displaystyle
f(x, n) = \prod_{k=0}^{n-1} \frac{(x+12k+3)^4+324}{(x+12k-3)^4+324}\,$.
Taking each term in turn, let $\,y=x+12k\,$ then the fraction can be written as $\,\displaystyle\frac{(y+3)^4+324}{(y-3)^4+324}\,$. Looking to simplify the fraction, it turns out that $\,\gcd\left((y+3)^4+324, (y-3)^4+324\right)=y^2+9\,$ (n.b. by polynomial Euclidean algorithm, not using any Sophie Germain's insight). After canceling out the common factor of $\,y^2+9\,$, the term becomes:
$$\require{cancel}
\displaystyle\frac{(y+3)^4+324}{(y-3)^4+324} = \displaystyle\frac{\cancel{(y^2+9)}(y^2+12y+45)}{\cancel{(y^2+9)}(y^2-12y+45)}=\frac{(y+6)^2+9}{(y-6)^2+9}=\frac{\big(x+12k+6\big)^2+9}{\big(x+12(k-1)+6\big)^2+9}
$$
It follows that the product telescopes nicely:
$$
\prod_{k=0}^{n-1} \frac{\big(x+12k+6\big)^2+9}{\big(x+12(k-1)+6\big)^2+9} \;=\; \frac{\big(x+12(n-1)+6\big)^2+9}{\big(x+12(0-1)+6\big)^2+9} \;=\; \frac{(x+12n-6)^2+9}{(x-6)^2+9}
$$
So in the end $\,\displaystyle f(7,5) = \frac{(7+5 \cdot 12 - 6)^2+9}{(7-6)^2+9} = \frac{61^2+9}{10}=\frac{3721+9}{10}=373\,$.
