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

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 $0$).
