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

This answer assumes that you want to find the minimum number of the real roots of $p(x)$.
Since $x=1$ is a root of $p(x)$, we have
$$p(1)=0\iff 2a+2b+3=0\iff b=\frac{-2a-3}{2}$$
from which we can write
$$p(x)=(x-1)^2\left(x^4+(a+2)x^3+\left(a+\frac 32\right)x^2+(a+2)x+1\right)$$
So, we see that the number of the real roots of $p(x)$ is equal to or more than $2$.
By the way, for $a=-2$, we have
$$p(x)=(x-1)^2\left(x^4-\frac 12x^2+1\right)=(x-1)^2\left(\left(x^2-\frac 14\right)^2+\frac{15}{16}\right)$$
of which $x=-1$ is not a root.
The number of the real roots of $p(x)$ for $a=-2$ is $2$.
Therefore, the minimum number of the real roots of $p(x)$ is $\color{red}{2}$.
