Solve $x^3(x^2+1)^2(x-1)(x+1)>0$ 
I need to solve this inequality:
$$x^3(x^2+1)^2(x-1)(x+1)>0$$

Note: I haven't learnt imaginary numbers. Does the $(x^2+1)^2$ affect the inequality?
Thanks
 A: Hint:  You have $(x^2+1)^2\gt0$ over the reals, therefore $$x^3(x^2+1)^2(x-1)(x+1)\gt0\iff x^3(x-1)(x+1)\gt0.$$
A: Firs of all, we have $x\not =0$.
Then, $x^2(x^2+1)^2\gt 0$, so the inequality is equivalent to
$$x(x-1)(x+1)\gt 0\iff x(x^2-1)\gt 0$$$$\iff "x\gt 0\ \text{and}\ x^2-1\gt 0"\ \ \text{or}\ \ "x\lt 0\ \text{and}\ x^2-1\lt 0"$$
$$\iff x\gt 1\ \ \text{or}\ \ -1\lt x\lt 0.$$
A: The term $(x^2+1)^2$ does not affect the inequality, since when you square a real number you always get a positive number, and $x^2+1$ is a real number, in your case. So what remains to look at is when
$$x^3(x-1)(x+1)>0$$
Now you have three critical point, if you want: $x_1=-1,$ $x_2=0$, $x_3=1$.
You need to look at what sign the function $f(x) := x^3(x-1)(x+1)$ has in each interval. For instance
$$\begin{array}{|c|c|c|c|c|}
\hline x\in&(-\infty,-1)&(-1,0)&(0,1)&(1,\infty)\\\hline \implies\text{signum $f(x)$} &-\cdot -\cdot -=-&-\cdot-\cdot+=+&+\cdot -\cdot +=-&+\cdot + \cdot +=+\\\hline\end{array}$$
Where for $x\in (-\infty,-1)$ we have for instance $x^3 <0$, $(x-1)<0$, $(x+1)<0$. (similarly for the other cases)
So you have that 
$$x^3 (x^2+1)^2(x-1)(x+1)>0 \iff x\in (-1,0)\cup (1,\infty)$$
($-1, 0, 1$ are not included since there you have $f(x)=0$ which is not greater than $0$).
A: IF you have not learned about complex numbers, I'll assume that $x \in \mathbb{R}$. Then, you can simplify the inequality as:
$$x^3(x-1)(x+1)>0$$ because $(x^2+1)^2>0\,\,\,\forall x\in\mathbb{R}$.
Now, the easiest way to solve this is searching for the critical points -where the expression on the LHS equals $0$- so we analize the intervals in between:
a) Search for critical points:
$$x^3(x-1)(x+1)>0\Rightarrow x^3=0 \vee (x-1)=0\vee (x+1)=0 $$
$$x^3=0\Rightarrow x=0$$
$$x-1=0\Rightarrow x=1$$
$$x+1=0\Rightarrow x=-1$$
b) Make a table with the real line, and these critical points marked. Then, analyze each expression that conforms the inequality, and anaalyze where they're positive or negative:
$$\begin{array}{c|ccccccc|c}
& &-1& &0& &1&&\\\hline
x^3&-&|&-&|&+&|&+&x^3>0\text{ when }x>0\\
x-1&-&|&-&|&-&|&+&x-1>0\text{ when }x>1\\
x+1&-&|&+&|&+&|&+&x+1>0\text{ when }x>-1\\\hline
x^3(x-1)(x+1)&-&|&+&|&-&|&+&\text{result of multiplying}\\
\end{array}$$
So, the solution is the union of all the cases where the expression is positive: 
$$x\in]-1,0[\,\cup\,]1,\infty[$$
