Ккто поможет обязательно поблагодарю .

Решите задачу:

$\frac{2(1-b^{2})}{4(b^{2}+2b+1)}= \frac{2(1-b)(1+b)}{4(b+1)^{2}}= \frac{1-b}{2(b+1)}$
$\frac{2(b^{2}-4b+4)}{4(4-b^{2})}= \frac{2(b-2)^{2}}{4(2-b)(2+b)}=\frac{2(2-b)^{2}}{4(2-b)(2+b)}=\frac{2-b}{2(2+b)}$
$\frac{x^{3}-3x^{2}+2x-6}{x^{3}-27}= \frac{x^{2}(x-3)+2(x-3)}{(x-3)(x^{2}+3x+9)}= \frac{(x^{2}+2)(x-3)}{(x-3)(x^{2}+3x+9)}= \frac{x^{2}+2}{x^{2}+3x+9}$
$\frac{x^{3}-8}{x^{3}-2x^{2}+x-2}= \frac{(x-2)(x^{2}+2x+4)}{x^{2}(x-2)+1(x-2)}= \frac{(x-2)(x^{2}+2x+4)}{(x^{2}+1)(x-2)}= \frac{(x^{2}+2x+4)}{x^{2}+1}$