Solve the inequation for $x$ Solve for $x$ :
$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $
I tried cases like : $ x-1 \le 0 $ , $ x \le 0 $ , $ x+1 \le 0 $

Well, let's see if we can simplify your expression first:
$$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $$
$$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$
Think, first, what qualifies $x$ to be $\le 0$? If $x$ is $0$ or a negative number, correct?
Well, our expression can be $0$ if $x = -1, 1$
Thus, now lets look when our expression yields a negative number.
Therefore, look at the powers and see which ones could be negative:
$$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$
We can see that $(x^2 - 1)^{2005}$ can only be negative, since the other $2$ terms have even powers, meaning they will only be positive.
Therefore, when is $(x^2 - 1)^{2005} \le 0$?
Well,
$$x^2 - 1 \le 0$$
$$x^2 \le 1$$
$$-1 \le x \le 1$$
And thus we arrive at our answer.
