Question

Упростите выражение (a+1)(a^2+1)(a^4+1)(a^8+1)(a^16+1)(a^32+1) 

Должно получиться (a^64 - 1)(a-1) 

 

 

Answer 1

$\frac{(a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}= \. \frac{(a^2-1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1} \frac{(a^4-1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}= \frac{(a^8-1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}= \frac{(a^{16}-1)(a^{16}+1)(a^{32}+1)}{a-1}= \frac{(a^{32}-1)(a^{32}+1)}{a-1}=frac{a^{64-1}}{a-1}$

Answer 2

$(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)= \ =frac{(a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}= \ =frac{(a^2-1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1} = \ =frac{(a^4-1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}= \ =frac{(a^8-1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}= \ =frac{(a^{16}-1)(a^{16}+1)(a^{32}+1)}{a-1}= \ =frac{(a^{32}-1)(a^{32}+1)}{a-1}=frac{a^{64-1}}{a-1}$