# Roots of a sixth degree polynomial

I have this question:

The polynomial $f(x) = x^6 - ax^4 - ax^2 +1$ has $(x-p)$ as a factor, where $a,p$ are real numbers. Show that $a = p^2 + p^{-2} - 1$

Here's my attempt:

Let $u = x^2 \implies f(x) = u^3 - au^2 - au + 1$

Let $\alpha, \beta, \gamma$ be the roots of this polynomial.

Since $(x-p)$ is a factor of the original polynomial, it follows that $\alpha = p^2$.

Also,

$\alpha + \beta + \gamma = a \implies \beta + \gamma = a - p^2$

$\alpha \beta \gamma = -1 \implies \beta \gamma = -\frac{1}{p^2}$

$\alpha \beta + \beta \gamma + \gamma \alpha = -a$

$\therefore p^2(\beta + \gamma) + \beta \gamma = -a$

$p^2(a-p^2) - \frac{1}{p^2} = -a$

I then continued in this vein for some time, trying to rearrange the expression to prove the result shown in the question. I couldn't do it. Am I going about this the right way, or am I completely wrong? If I'm right, where do I go from here?

The polynomial $f(x) = x^6 - ax^4 - ax^2 +1$ has $(x-p)$ as a factor iff $f(p)=0$.

Now $0=f(p)=p^6 - ap^4 - ap^2 +1$ and so $a(p^4+p^2)=p^6+1=(p^2+1) (p^4-p^2+1)$, because $u^3+1=(u+1)(u^2-u+1)$.

Cancelling $p^2+1$, which is never zero, we get $ap^2 = p^4-p^2+1$, which is equivalent to $a = p^2 + p^{-2} - 1$ because $p\ne 0$.

HINT:

$$p^6-ap^4-ap^2+1=0$$

$$p^6-ap^4-ap^2+1=(p^2)^3+1-ap^2(p^2+1)=(p^2+1)\{(p^4-p^2+1)-ap^2\}$$