<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] My SOS is ugly.
$$x^{12}-x^9+x^4-x+1$$
$$={\frac { ( 8x^6-4x^3-1)^2}{64}}+{\frac { \left( 80x^
2-5x-72 \right) ^{2}}{6400}}+{\frac {2299}{1280} \left( x-{\frac {
712}{2299}} \right) ^{2}}+{\frac {7741}{3678400}}.$$
Remark. From Mr. Mike solution we can get
$$x^{12}-x^9+x^4-x+1={\frac {{x}^{13} \left( {x}^{2}+x+1 \right)  \left( {x}^{8}+1 \right) 
+ \left( {x}^{2}+x+1 \right)  \left( {x}^{6}+{x}^{3}-x+1 \right) }{
 \left( {x}^{2}+x+1 \right)  \left( x+1 \right)  \left( {x}^{2}+1
 \right)  \left( {x}^{2}-x+1 \right)  \left( {x}^{4}-{x}^{2}+1
 \right) }}$$
For $x>0,$ it's clearly true!
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