<s>[INST] Solving $\log_{1/3} \left(x^3 - 1\right) > \log_{1/3} \left(|3x^2 - 3x|\right)$ I tried over and over again but I simply can't solve:
$$\log_{1/3} \left(x^3 - 1\right) > \log_{1/3} \left(|3x^2 - 3x|\right)$$
So, if someone could please explain to me how to solve this step by step, I would be very greatful. By the way, the solution for this problem is $x \in \varnothing$.

Edit
This is what I tried to do:
$$\log_\frac{1}{3} \left(x^3 - 1\right) > \log_\frac{1}{3} \left(|3x^2 - 3x|\right) $$
to find the domain
$$\begin{cases} 
x^3-1>0 \implies x^3>1 \implies x>\sqrt[3] 1 \implies x>1 \\
3x^2 - 3x>0 \implies \ldots \implies x\in(-\infty,0)\cup(1,\infty)\\
-3x^2 + 3x<0 \implies \ldots \implies x\in(-\infty,0)\cup(1,\infty)
\end{cases}$$
then domain is $x>1$.
We need to solve
$$x^3 - 1 <|3x^2 - 3x| \implies 
\begin{cases} 
x^3 - 1 <3x^2 - 3x \quad (1)\\
x^3 - 1 <-3x^2 + 3x \quad (2)
\end{cases}$$
from $(1)$ we otain
$$x^3 -3x^2 + 3x-1<0 \implies x<1$$
from $(2)$ we otain
$$x^3 +3x^2 - 3x-1<0  \implies x\in(-\infty,-2-\sqrt 3)\cup(-2-\sqrt 3,1)$$
 [/INST] We have that $\log_{\frac 13}x$ function is strictly decreasing then
$$\log_\frac{1}{3} \left(x^3 - 1\right) > \log_\frac{1}{3} \left(|3x^2 - 3x|\right) \iff x^3 - 1 <|3x^2 - 3x|$$
under the conditions

*

*$x^3 - 1>0 \iff x>1$

*$|3x^2 - 3x| \neq 0 \iff x\neq 0 \quad x \neq 1$
that is $x>1$ therefore
$$x^3 - 1 <|3x^2 - 3x| \iff x^3 - 1 <3x^2 - 3x \iff x^3-3x^2+3x - 1=(x-1)^3<0$$
</s>