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?
 A: Your condition isn't quite right.  When $x = 1.1$, $x^3−x−1 = 1.331 - 1.1 - 1 < 0$.
From $\log(x^3 - x - 1) < \log(x^3 + x - 1)$, subtract one side from the other, then use a logarithm property to get one logarithm.  \begin{align*}
0 &< \log(x^3 + x - 1) - \log(x^3 - x - 1)  \\
0 &< \log \left( \frac{x^3 + x - 1}{x^3 - x - 1} \right)
\end{align*}
Apply $f(y) = 10^y$ to both sides to obtain
\begin{align*}
1 &< \frac{x^3 + x - 1}{x^3 - x - 1}  \text{,}
\end{align*}
which you demonstrated you knew how to handle in your question.  (Don't forget, when you go from $a < \frac{b}{c}$ to $ac < b$ you may have inadvertently multiplied by zero for some values of $x$ that you aren't currently thinking about (yielding $0 < 0$, which is false).  You should check what happens in those cases.  For instance, here, what happens when $x^3 - x - 1 = 0$ in the original equation?)
A: 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:
$$(1,+\infty)\setminus\{x|x^3-x-1=0\}.$$
A: Better not get rid of the squares -- they're your friend. So we have $$\log{(x^3-x-1)^2}< \log{(x^3+x-1)^2}.$$ Since the logarithm is monotonic, this implies $$(x^3-x-1)^2< (x^3+x-1)^2,$$ or that $$(x^3-x-1)^2- (x^3+x-1)^2<0.$$ This is easily factored to give $$(x^3-x-1+x^3+x-1)(x^3-x-1-x^3-x+1)<0,$$ or $$(2x^3-2)(-2x)<0,$$ or more simply $$x(x^3-1)=x(x-1)(x^2+x+1)>0,$$ which implies $$x(x-1)>0$$ since $x^2+x+1$ is always positive.
I believe you can now take it from here.

Well, you also have to factor in that the original inequality makes sense only for $x^3+x-1>0$ and $x^3-x-1\ne 0.$
