How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$       (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumped me, but then I realised I can just add the inequalities to get:
$2x^3 -2> 0$
$x^3-1>0$
$x^3>1 \Rightarrow x \in (1, + \infty)$
So that is our condition.
Going back to (*) and raising the inequality to the power of $10$, I got:
$x^3-x-1 < x^3+ x - 1$
$2x>0 \Rightarrow x \in(0, +\infty)$
So, if we also consider the condition, we have
$x \in (1, + \infty) \cap (0, + \infty)$
So $x \in (1, + \infty)$.
The problem with this answer is that it is wrong. My textbook lists the following possible answers:
A. $\mathbb{R}$
B. $(0, + \infty)$
C. $(1, + \infty)$
D. $(0, 1)$
E. Other answer
So I got answer C, but after I checked the back of the book I found that the correct answer should be E. So, what did I do wrong and what is that other answer?

The domain it's $$x^3-x-1\neq0$$ and $$x^3+x-1>0.$$ 
Now, since $$\ln(x^3-x-1)^2=2\ln|x^3-x-1|,$$ we need to solve
$$|x^3-x-1|<x^3+x-1$$ or
$$-x^3-x+1<x^3-x-1<x^3+x-1.$$
The left inequality gives $x>1$ and the right inequality gives $x>0,$ which gives the answer:
$$(1,+\infty)\setminus\{x|x^3-x-1=0\}.$$
