315*. Find all roots of the equation

$$
8 x\left(2 x^{2}-1\right)\left(8 x^{4}-8 x^{2}+1\right)=1
$$

satisfying the condition $0<x<1$.
$\triangle$ Let's introduce a trigonometric substitution $x=\cos \alpha$, where $\alpha \in\left(0 ; \frac{\pi}{2}\right)$. Then

$\cos 2 \alpha=2 \cos ^{2} \alpha-1=2 x^{2}-1$,

$\cos 4 \alpha=2 \cos ^{2} 2 \alpha-1=2\left(2 x^{2}-1\right)^{2}-1=8 x^{4}-8 x^{2}+1$.

The original equation transforms into a trigonometric equation:

$$
8 \cos \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha=1
$$

Multiply it by $\sin \alpha$:

$\sin 8 \alpha=\sin \alpha, \quad \sin 8 \alpha-\sin \alpha=0, \quad 2 \cos \frac{9 \alpha}{2} \sin \frac{7 \alpha}{2}=0$.

Set $\cos \frac{9 \alpha}{2}$ to zero:

$$
\frac{9 \alpha}{2}=\frac{\pi}{2}, \alpha=\frac{\pi}{9} ; \quad \frac{9 \alpha}{2}=\frac{3 \pi}{2}, \alpha=\frac{\pi}{3} ; \quad \frac{9 \alpha}{2}=\frac{5 \pi}{2}, \alpha=\frac{5 \pi}{9}
$$

The last value of $\alpha$ does not belong to the interval $\left(0 ; \frac{\pi}{2}\right)$, so it should be discarded.

Similarly, solve the equation $\sin \frac{7 \alpha}{2}=0$:

$$
\frac{7 \alpha}{2}=\pi, \alpha=\frac{2 \pi}{7} ; \quad \frac{7 \alpha}{2}=2 \pi, \alpha=\frac{4 \pi}{7}
$$

The root $\alpha=\frac{4 \pi}{7}$ is also extraneous.

Answer: $\cos \pi / 9, \cos \pi / 3, \cos 2 \pi / 7$.