86. Using the equality $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$, solve the equation

$$
x^{3}-1=0
$$

Applying the obtained formulas, find the three values of $\sqrt[3]{1}$ in 103 -arithmetic.
86. From the equality $(x-1)\left(x^{2}+x+1\right)=0$ it follows that

$$
x-1=0
$$

or

$$
x^{2}+x+1=0
$$

From the first equation we find

$$
x_{1}=1
$$

from the second

$$
x_{2,3}=\frac{-1 \pm \sqrt{-3}}{2}
$$

In 103-arithmetic $-3=100$ and $\sqrt{-3}= \pm 10$. Substituting these values of $\sqrt{-3}$ into formula (1), we get

$$
x_{1}=1, \quad x_{2}=56, \quad x_{3}=46
$$