You could prove it with just a couple of high-school trigonometric identities.
Let $\theta = \pi/7 $ and recognize
$$x_1=\cos \theta = -\cos 6\theta$$
$$x_2=\cos3\theta = -\cos 4\theta$$
$$x_3=\cos5\theta = -\cos 2\theta$$
(1). Evaluate their product by applying the identity $\sin 2x = 2\sin x \cos x$
$$ \cos2\theta \cos4\theta \cos6\theta = \frac{\sin 4\theta \cos 4\theta\cos 6\theta}{2\sin 2\theta} = \frac{\sin 8\theta \cos 8\theta }{4\sin 2\theta}= \frac{1}{8}$$
(2). Evaluate their sum by applying $\sin(x+y)+\sin(x-y)=2\sin x\cos y$,
$$ 2\sin 2\theta(\cos2\theta + \cos4\theta + \cos6\theta)$$
$$=\sin 4\theta + (\sin 6\theta - \sin 2\theta) + (\sin 8\theta - \sin 4\theta)$$
which after some cancellation leads to,
$$\cos2\theta + \cos4\theta + \cos6\theta = -\frac{1}{2}$$
(3). Evaluate the sum of their cross products by applying $\cos(x+y)+\cos(x-y)=2\cos x\cos y$,
$$\cos4\theta \cos6\theta + \cos6\theta \cos2\theta + \cos2\theta \cos4\theta $$
$$= \cos2\theta + \cos4\theta + \cos6\theta = -\frac{1}{2}$$
So, in terms of $x_1$, $x_2$ and $x_3$,
$$x_1+x_2+x_3=\frac{1}{2}$$
$$x_1x_2+x_2x_3+x_3x_1=-\frac{1}{2}$$
$$x_1x_2x_3=-\frac{1}{8}$$
And
$$(x-x_1)(x-x_2)(x-x_3)=x^3 - \frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{8}$$
Therefore, $x_1=\cos \theta$, $x_2=\cos 3\theta$ and $ x_3=\cos 5\theta$ are the roots of
$$x^3-\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{8}=0$$
or,
$$8x^3-4x^2-4x+1= 0$$
