<s>[INST] How to calculate $3*5*17*257-3^{16}$ using factorization formulas? look at this:
$2*4*10*82*6562-3^{16}$
It's easy to calculate it with elementary arithmetic.
but how to calculate it using factorization formulas?
 [/INST] Hint $\ $ It is a radix $\,3\,$ analog of this radix $\,10\,$ (decimal) computation
$\qquad\qquad\qquad\qquad  \underbrace{9\cdot 11}\cdot 101\cdot 10001 - 10^8 $
$\qquad\qquad\qquad\qquad\quad \underbrace{99 \cdot 101}$
$\qquad\qquad\qquad\qquad\qquad\ \  \underbrace{9999\cdot 10001}$
$\qquad\qquad\qquad\qquad\qquad\quad\ \ 99999999 - 10^8\, =\, -1$
Remark $\ $ At the heart of the matter is the following telescoping product  (in your case $\,x=3).\,$ Notice that the same-colored terms all (diagonally) cancel out of the product.
$\qquad\qquad\, \displaystyle (x-1)(x+1)(x^2\!+1)(x^4\!+1)\quad\ \ \cdots\quad\ \ \ (x^{2^{\rm N}}\!+1)$
$\qquad\ \ \ = \ \displaystyle  \frac{\color{#0a0}{x-1}}{\color{#90f}1} \frac{\color{brown}{x^2-1}}{\color{#0a0}{x-1}}\frac{\color{royalblue}{x^4-1}}{\color{brown}{x^2-1}}\frac{\phantom{f(3)}}{\color{royalblue}{x^4-1}}\, \cdots\,  \frac{\color{#c00}{x^{2^{\rm N}}\!-1}}{\phantom{f(b)}}\frac{x^{2^{\rm N+1}}\!-1}{\color{#c00}{x^{\rm 2^N}\!-1}} \,=\,  \frac{x^{2^{\rm N+1}}-1}{\color{#90f}1} $
For further telescopic intuition see my many posts on multiplicative telescopy.
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