Question

(a^6+a^4+a^2+1)/(a^3+a^2+a+1) Нужно сократить дробь

Answer 1

$\displaystyle \frac{a^6+a^4+a^2+1}{a^3+a^2+a+1}=\frac{a^4(a^2+1)+a^2+1}{a^2(a+1)+a+1}=\frac{(a^4+1)(a^2+1)}{(a^2+1)(a+1)}=\frac{a^4+1}{a+1}$

Answer 2

$\displaystyle \frac{a^6+a^4+a^2+1}{a^3+a^2+a+1} =\frac{a^4*a^2+a^4+a^2+1}{a^2*a+a^2+a+1}=\\\\=\frac{a^4(a^2+1)+a^2+1}{a^2(a+1)+a+1} =\frac{a^4(a^2+1)+(a^2+1)}{a^2(a+1)+(a+1)} =\\\\=\frac{(a^2+1)(a^4+1)}{(a+1)(a^2+1)} =\frac{(a^2+1)(a^4+1)}{(a^2+1)(a+1)} =\\\\=\frac{a^4+1}{a+1}$