Помогите пожалуйста

Решите задачу:

$1)\; \; \frac{x^2-y^2}{y^3-x^3}= \frac{(x-y)(x+y)}{-(x-y)(x^2+xy+y^3)} =-\frac{x+y}{x^2+xy+y^2y} \\\\2)\; \; \frac{2a^4b^3+8a^3b^4+8a^2b^5}{5a^2b^2+10ab^3} = \frac{2a^2b^3(a^2+4ab+4b^2)}{5ab^2(a+2b)} = \frac{2a^2b^3(a+2b)^2}{5ab^2(a+2b)} = \\\\=\frac{2ab(a+2b)}{5} =0,4ab\cdot (a+2b)$

$3)\; \; a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc=\\\\=a(b^2+2bc+c^2)+b(c^2+2ac+a^2)+c(a^2+2ab+b^2)-4abc=\\\\=ab^2+2abc+ac^2+bc^2+2abc+a^2b+a^2c+2abc+b^2c-4abc=\\\\=ab^2+a^2b+ac^2+a^2c+bc^2+b^2c+2abc\; ;\\\\\\(a+b)(b+c)(c+a)=(ab+ac+b^2+bc)(c+a)=\\\\=abc+a^2b+ac^2+a^2c+b^2c+ab^2+bc^2+bac=\\\\=a^2b+ab^2+ac^2+a^2c+bc^2+b^2c+2abc\; ;\\\\\Rightarrow \\\\a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc=(a+b)(b+c)(c+a)$