## Task B-2.2.

Solve the equation $\left(16^{-x}-2\right)^{3}+\left(4^{-x}-4\right)^{3}=\left(16^{-x}+4^{-x}-6\right)^{3}$.
## Solution.

Let's introduce the substitution $16^{-x}-2=a$ and $4^{-x}-4=b$. Then the given equation becomes $a^{3}+b^{3}=(a+b)^{3}$. It follows that:

$$
\begin{aligned}
& a^{3}+b^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\
& 3 a b(a+b)=0 \\
& a=0 \text { or } b=0 \text { or } a+b=0
\end{aligned}
$$

If $a=0$, then $16^{-x}-2=0$, which gives $x=-\frac{1}{4}$.

If $b=0$, then $4^{-x}-4=0$, which gives $x=-1$.

If $a+b=0$, then $16^{-x}+4^{-x}-6=0$. Let's introduce the substitution $t=4^{-x}$.

Then we have $t^{2}+t-6=0$, which gives $t=2$ or $t=-3$. Since $t>0$, it must be $t=2$, which gives $x=-\frac{1}{2}$.

Therefore, the set of solutions to the given equation is $\left\{-1,-\frac{1}{2},-\frac{1}{4}\right\}$.