Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$ 
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.

I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing

The sum of square form
$$2(x^{12}-x^9+x^4-x+1)$$
$$=x^6(x^3-1)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{4}\right)^2+(x-1)^2+\frac{5}{8}>0.$$
