<s>[INST] then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$ 
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is 
Attempt: $x=0$ in not a root of $p(x)=0.$ 
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)+a\left(x+\frac{1}{x}\right)^2-2+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle t^3+at^2+(b-3)t+1=0,$ where $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $|t|\geq 2$
Could some help me to solve it, thanks
 [/INST] We assume you want real roots (complex roots are 6 exactly).
$x=1$ is one of the root of the main equation so $t=2$ is the root of the second, write
$$\displaystyle t^3+at^2+(b-3)t+1-2a=0~~;~~~t\neq0~~(x\neq-1)$$
$$(t-2)(t^2+mt+n)=0~~;~~~t\neq0$$
this concludes that $n=a-\dfrac12$ and $m=a+2$.
$t\neq0$ says $n=a-\dfrac12\neq0$ and for $t^2+mt+n=t^2+(a+2)t+(a-\dfrac12)=0$ we see
$$t_1=\dfrac{-(a+2)+\sqrt{a^2+6}}{2}~~~,~~~t_2=\dfrac{-(a+2)-\sqrt{a^2+6}}{2}$$
if $a>-\dfrac14$ then $t_2<-2$ thus $t+\dfrac1t=t_2$ has two roots.
if $a<-\dfrac52$ then $t_1>2$ thus $t+\dfrac1t=t_1$ has two roots.
for $-\dfrac52<a<-\dfrac14$ is not determined if the eqution has other root(s) or not.
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