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
 A: 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 $$x^{12}-x^9+x^4-x+1>0.$$
A: Hint: Break it into cases:
If $x \ge 1$, then $x^{12} \ge x^9$ and $x^4 \ge x$.
If $x \le 0$, then $x^{12} \ge 0$ and $x^9 \le 0$ and $x^4 \ge 0$ and $x \le 0$.
If $0 < x < 1$, then $x^{12} > 0$ and $x^4 > x^9$ and $1 > x$.
Can you finish each of these cases?
A: The sum of square form
$$2(x^{12}-x^9+x^4-x+1)$$
$$=x^6(x^3-1)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{4}\right)^2+(x-1)^2+\frac{5}{8}>0.$$
A: Another way.
For $x\leq0$ it's obvious.
But for $x>0$  by AM-GM we obtain:
$$x^{12}-x^9+x^4-x+1=x^{12}-x^9-x^4+x+2x^4-2x+1=$$
$$=x(x^8-1)(x^3-1)+2x^4+3\cdot\frac{1}{3}-2x\geq$$
$$\geq4\sqrt[4]{2x^4\cdot\left(\frac{1}{3}\right)^3}-2x=2\left(\sqrt[4]{\frac{32}{27}}-1\right)x>0.$$
A: If $x\leq0$  and $x\geq1$ it is obvius.
If $0<x<1$ take $x^4>x^9$ and $1>x$
A: 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!
