(x^2+1)(x^3-1)(x^4-16)<0

Решите задачу:

$(x^2+1)(x^3-1)(x^4-16)\ \textless \ 0\\\\x^2+1\ \textgreater \ 0\; \; \to \; \; \; (x^3-1)(x^4-16)\ \textless \ 0\\\\(x-1)\underbrace{(x^2+x+1)}_{\ \textgreater \ 0}(x^2-4)\underbrace {(x^2+4)}_{\ \textgreater \ 0}<0\\\\(x-1)(x-2)(x+2)<0\\\\---(-2)+++(1)---(2)+++\\\\x\in (-\infty ,-2)\cup (1,2)\\\\P.S.\; \; \; x^2+x+1\ \textgreater \ 0\; ,\; tak\; kak\; \; D=1-4=-3\ \textless \ 0$