Domain of the Function Square Root of 12th Degree Polynomial Find the Domain of $$f(x)=\frac{1}{\sqrt{x^{12}-x^9+x^4-x+1}}$$
My Try: The Domain is given by
$$x^{12}-x^9+x^4-x+1 \gt 0$$ $\implies$
$$x(x-1)(x^2+x+1)(x^8+1)+1 \gt 0$$
Please help me how to proceed further..

If $x$ is out of the interval $[0,1]$ all the facotrs in $$x(x-1)(x^2+x+1)(x^8+1)+1$$
are positive so we have $$x(x-1)(x^2+x+1)(x^8+1)+1\gt 0$$
But for $x\in [0,1],$ $$x(x-1)\geq -\frac{1}{4}$$
So
$$x(x-1)(x^2+x+1)(x^8+1)+1\geq -\frac{1}{4}\cdot (x^2+x+1)(x^8+1)+1\geq -\frac{1}{4}\cdot\frac{3}{4}.1+1=\frac{13}{16}$$
i.e., for all $x\in \mathbb R$
$$x(x-1)(x^2+x+1)(x^8+1)+1\gt 0$$
