Решение по пунктам Очень нужно У самой мозгов не хватает

Решите задачу:

$1) \frac{a^3 +b^3+3ab(a+b)}{(a+b)^3}*2= \frac{(a+b)^3}{(a+b)^3} *2=2\\ 2) \frac{(a^2-ab+b^2)(a^2+2ab+b^2)(a+b)}{(a+b)^2(a^3+b^3)}= \frac{(a+b)^2(a^3+b^3)}{(a+b)^2(a^3+b^3)} =1\\$
$3) \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)}= \\ =\frac{a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3}{(a-b)(b-c)(c-a)} =\\ = \frac{3c(a+b)(a-b)-3ab(a-b)-3c^2(a-b)}{(a-b)(b-c)(c-a)} =\\ = \frac{3(a-b)(ac+bc-ab-c^2)}{(a-b)(b-c)(c-a)} =\\ = \frac{3(c(b-c)-a(b-c))}{(b-c)(c-a)} =\\ = \frac{3(b-c)(c-a)}{(b-c)(c-a)} =3$
$4) \frac{(x+y+z)^3-x^3-y^3-z^3}{(x+y)(y+z)(z+x)}*9=\\ = \frac{x^3+y^3+z^3+3x^2y+3xy^2+3(x+y)^2z+3(x+y)z^2-x^3-y^3-z^3}{(x+y)(y+z)(z+x)}*9=\\ = \frac{3xy(x+y)+3z(x+y)^2+3z^2(x+y)}{(x+y)(y+z)(z+x)} *9=\\ = \frac{3(x+y)(xy+zx+zy+z^2)}{(x+y)(y+z)(z+x)} *9=\\ = \frac{27(x(y+z)+z(y+z))}{(y+z)(z+x)} =\\ = \frac{27(y+z)(z+x)}{(y+z)(z+x)} =27$
$5) \frac{a^3+b^3+c^3-3abc-a^3-ab^2-ac^2+a^2b+abc+a^2c}{a^2+b^2+c^2-ab-ac-bc}+ \\ + \frac{-b^3-a^2b-bc^2+ab^2+b^2c+abc-c^3-a^2c-b^2c+abc+bc^2+ac^2}{a^2+b^2+c^2-ab-ac-bc} -1 =\\ = 0-1 = -1$