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

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?
