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?
 A: $$2\times 4\times 10\times 82\times 6562-3^{16} $$ $$=(3^1-1)(3^1+1)(3^2+1)(3^4+1)(3^8+1)-3^{16}$$ $$=(3^2-1)(3^2+1)(3^4+1)(3^8+1)-3^{16}$$ $$=(3^4-1)(3^4+1)(3^8+1)-3^{16}$$ $$=(3^8-1)(3^8+1)-3^{16}$$ $$=(3^{16}-1)-3^{16}$$ $$=-1$$ 
A: 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.
A: Hints: $$3 \cdot 5 \cdot 17 \cdot 257 = (2-1)(2+1)(2^2+1)(2^4+1)(2^8+1) = (2^2-1)(2^2+1)(2^4+1)(2^8+1) = \dots$$
$$2 \cdot 4 \cdot 10 \cdot 82 \cdot 6562 = (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)=\dots$$
A: First of all, the answer is quite a big number. So I don't think the exact value matters. I think you are asking how can you "compactify" your answer. I am assuming that your heading is your question. Write it as $(2+1)(2^2+1)(2^4+1)(2^8+1)-3^{16}$. Multiply the big 4 terms in brackets by $(2-1)$(its nothing but $1$). This starts your process of chain reaction. Multiply it by first term in bracket and you get $(2+1)(2-1)=2^2-1$, which in turn multiplies by the next term, then the next.... At last you get $2^16-1$. And therefore your answer is $$2^{16}-3^{16}-1$$. 
