(x^3-x^2+x-1)/(x+8)<0
[x²(x-1)+(x-1)]/(x+8)<0<br>(x²+1)(x-1)/(x+8)<0<br>x²+1>0 при любом х⇒(x-1)/(x+8)<0<br>x=1 x=-8 x∈(-8;1)