Question

1. (x2-1)/(1-x3)-x2/(x2+x+1) 2. (x4-y4)/(x3+yx2-xy2-y3)-2y 3. a/(a2-3a)+(a-8)/a+(7a2-3a-72)/(a3-9a)

Answer 1

1. 
$\frac{x^2-1}{1-x3}-\frac{x^2}{x^2+x+1}=\frac{x^2-1}{(1-x)(1+x+x^2)}-\frac{x^2}{x^2+x+1}=\frac{x^2-1-x^2+x^3}{(1-x)(x^2+x+1)}=\\=\frac{x^3-1}{1-x^3}=-1$

2.
$\frac{x^4-y^4}{x^3+yx^2-xy^2-y^3}-2y=\frac{(x^2-y^2)(x^2+y^2)}{x^2(x+y)-y^2(x+y)}-2y=\frac{(x^2-y^2)(x^2+y^2)}{(x+y)(x^2-y^2)}-2y=\\=\frac{x^2+y^2}{x+y}-2y=\frac{x^2+y^2-2xy-2y^2}{x+y}=\frac{x^2-2xy-y^2}{x+y}$

3.
$\frac{a}{a^2-3a}+\frac{a-8}{a}+\frac{7a^2-3a-72}{a^3-9a}=\frac{a}{a(a-3)}+\frac{a-8}{a}+\frac{7a^2-3a-72}{a(a-3)(a+3)}=\\= \frac{a(a+3)+(a-8)(a^2-9)+7a^2-3a-72}{a(a-3)(a+3)}=\\= \frac{a^2+3a+a^3-9a-8a^2+72+7a^2-3a-72}{a(a-3)(a+3)}=\frac{a^3-9a}{a(a-3)(a+3)}=\frac{a(a-3)(a+3)}{a(a-3)(a+3)}=1$