<s>[INST] 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
 [/INST] For $x\geq1$ we obtain:
$$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1>0.$$
For $0<x<1$ we have:
$$x^{12}-x^9+x^4-x+1=(1-x)+x^4(1-x^5)+x^{12}>0.$$
For $x\leq0$ it's obvious that $$x^{12}-x^9+x^4-x+1>0.$$
</s>