Roots of unity plays a role here?

Without using roots of unity, we try to solve the problem...

Lemma: If $x^n = x^k = 1$ , where $gcd(n,k) = 1$ , then $x = 1$

Proof: Since $gcd(n,k) = 1$ , by Bezout's identity, we can find integers $a,b$ such that $an- bk = 1$ .

Therefore, $x^{an} = x^{bk} = 1 \implies x^{an} - x^{bk} = 0 \implies x^{bk}(x-1) = 0 \implies x=1$ , since $x \neq 0$ .

Thus the proof is complete. This lemma implies that we cannot have $x^7 = 1$ since this means that one other number in the list is equal to 1, and this makes all the numbers equal to 1, since $7$ is coprime with $6,8,9$ . If any of these numbers were equal to 1, then by our lemma, $x = 1$ therefore all of these are equal to 1. Similarly, we cannot have both $x^8 =1, x^9 = 1$ by the same reason. Therefore, by the pigeonhole principle, $\boxed{x^6 = 1}$ .

For $x^k$ , where $k = 6,7,8,9$ , to be equal to 1, $x^k$ must be equal to $e^{2n\pi i} = 1$ , where $n$ is a natural number. Note that $x = e^{2n \pi i}$ is not acceptable, because all four $x^k = 1$ . The possible $x$ for each $k$ are as follows:

$\begin{cases} k = 6 & \implies \begin{cases} {\color{#3D99F6}x = e^{\frac 23 \pi i}} & \implies x^6 = e^{2\pi i} = 1 \\ {\color{#3D99F6}x = e^{\pi i}} & \implies x^6 = e^{6 \pi i} = 1 \end{cases} \\ k = 7 & \implies x = e^{\frac 27 \pi i} \quad \ \ \ \implies x^7 = e^{2\pi i} = 1 \\ k = 8 & \implies \begin{cases} x = e^{\frac 14 \pi i} & \implies x^8 = e^{2\pi i} = 1 \\ x = e^{\frac 12 \pi i} & \implies x^8 = e^{4\pi i} = 1 \\ {\color{#3D99F6}x = e^{\pi i}} & \implies x^8 = e^{8 \pi i} = 1 \end{cases} \\ k = 9 & \implies \begin{cases} {\color{#3D99F6}x = e^{\frac 23 \pi i}} & \implies x^9 = e^{6\pi i} = 1 \\ x = e^{\frac 29 \pi i} & \implies x^9 = e^{2 \pi i} = 1 \end{cases} \end{cases}$

There are only two $x$ , when two of the four $x^k = 1$ : $\begin{cases} x = e^{\frac 23 \pi i} & \implies {\color{#3D99F6}x^6}=x^9 =1 \\ x = e^{\pi i} & \implies {\color{#3D99F6}x^6}=x^8 =1 \end{cases}$

In both cases $\boxed{x^6}$ appears.

There are only two ways to satisfy precisely two of the above answers = 1.

1) $x =$ one of the complex cube roots of $1$

$\rightarrow x^6 = 1$ and $x^9 = 1$

2) $x = -1$

$\rightarrow x^6 = 1$ and $x^8 = 1$

Either way $\boxed{x^6 = 1}$

Further explanation:

For x^7 = 1, there are 7 roots, which are all complex except for x=1. And none of the complex ones are roots of the other three equations. But, if x=1 then they all would equal 1.

If $x^8 = 1$ , then there are $8$ roots, $\pm1$ , $\pm i$ , and $\pm\frac{1}{\sqrt{2}} \pm\frac{1}{\sqrt{2}}i$ . The only ones of these which raised to the 6th power equal one are $\pm1$ . So, for $-1$ precisely two will work.

And if $x^9 = 1$ , then there are $9$ roots, $1$ and some complex ones. Only two of the complex roots will also satisfy $x^6 = 1$ , and none only satisfy $x^8 = 1$ or $x^7 =1$ .