limits - Evaluate $lim_{x to -infty} ln(-x^3+x)$

Evaluate $$\lim_{x \to -\infty} \ln(-x^3+x).$$

I was wondering if I can solve this limit in this way:
$$\lim_{x \to -\infty} \ln(-x^3+x)=\lim_{x \to -\infty} \ln\left[x^3\left(1+\frac{1}{x^2}\right)\right].$$

At this point, I just considered $\ln(x^3)$ because $1$ doesn't make any difference and $1/x^2$ tends towards $0.$
So, the result will be $0.$ And I found it because I know the graph of the logarithm of $x$ to the power of an odd number. So, my second question is, is it possible to understand what the result of $\lim_{x\to - \infty} \ln(x^3)$ algebraically without thinking of the graph?

Any suggestion and help will be appreciated.
Thank you in advance and have a good day :)

Answer

The correct way is

$$\lim_{x \to -\infty} \ln(-x^3+x)=\lim_{x \to -\infty} \ln\left[-x^3\left(1-\frac{1}{x^2}\right)\right]=\lim_{x \to -\infty} \left[\ln -x^3+\ln\left(1-\frac{1}{x^2}\right)\right]=\\=\lim_{x \to -\infty} \ln -x^3+\lim_{x \to -\infty} \ln\left(1-\frac{1}{x^2}\right)=+\infty+0=+\infty$$

indeed

$$\lim_{x \to -\infty} \ln -x^3=\lim_{x \to -\infty} 3\ln -x=\lim_{y \to +\infty} 3\ln y=+\infty$$