82. Let's get rid of the irrationality in the denominator:

$$
\frac{2}{1-\sqrt[3]{3}-\sqrt[3]{9}}
$$
Solution. Let $\frac{2}{1-\sqrt[3]{3}-\sqrt[3]{9}}=a+b \sqrt[3]{3}+c \sqrt[3]{9}$, then

$$
(1-\sqrt[3]{3}-\sqrt[3]{9})(a+b \sqrt[3]{3}+c \sqrt[3]{9})=2
$$

or

$$
(a-3 b-3 c)+(b-a-3 c) \sqrt[3]{3}+(c-a-b) \sqrt[3]{9}=2
$$

From this, $\left\{\begin{array}{l}a-3 b-3 c=2, \\ b-a-3 c=0, \\ c-a-b=0\end{array}\right.$

Solving the system, we have $a=\frac{1}{5}, b=-\frac{2}{5}$ and $c=-\frac{1}{5}$.

Therefore, $\frac{2}{1-\sqrt[3]{3}-\sqrt[3]{9}}=\frac{1}{5} \cdot(1-2 \sqrt[3]{3}-\sqrt[3]{9})$.

On the solution of a class of cubic equations

Let the cubic equation be given

$$
a_{1} x^{3}+b_{1} x^{2}+c_{1} x+d_{1}=0\left(a_{1} \neq 0\right)
$$

We bring this equation to the form

$$
x^{3}+a x^{2}+b x+c=0
$$

where $a=\frac{b_{1}}{a_{1}}, b=\frac{c_{1}}{a_{1}}, c=\frac{d_{1}}{a_{1}}$.

Let $x=y+m$ in equation (1). Then

$$
y^{3}+y^{2}(a+3 m)+y\left(3 m^{2}+2 a m+b\right)+m^{3}+a m^{2}+b m+c=0
$$

For equation (2) to be a binomial, the following condition must be satisfied:

$$
\left\{\begin{array}{l}
a+3 m=0 \\
b+2 a m+3 m^{2}=0
\end{array}\right.
$$

Solutions to this system: $m=-\frac{a}{3}, a^{2}=3 b$.

Thus, for any $c$ and for $a^{2}=3 b$, equation (1) by setting $x=y-\frac{a}{3}$ can be reduced to a binomial cubic equation.