Minima and polynomials?

This equation is symmetric so we can divide her by $x^3$ $(x\ne 0)$ so we get $x^3+1/x^3+3a(x^2+1/x^2)+(3a^2+3)(x+1/x)+6a+a^3=0$

and now replace $t=x+1/x$ ,so $t^2-2=x^2+1/x^2$ and $t^3-3t=x^3+1/x^3$

Now we have that $t^3-3t+3a(t^2-2)+(3a^2+3)t+6a+a^3=0$

So, $t^3+3at^2+3a^2t+a^3=0 i.e. (t+a)^3=0$ so $t=-a$ is triple solution. Now we have $x+1/x=-a$ ,so $x^2+ax+1=0$ and that means that $a^2-4\geq 0$ (discriminant $\geq 0$ for real solutions), so min $a =2$ .That means $1000a=2000$

We have:

$\quad\; x^6+3ax^5+(3a^2+3) x^4+6ax^3+(3a^2+3) x^2+3ax+1+(ax)^3=0$

$\Leftrightarrow (x^2+ax+1)^3=0$

$\Leftrightarrow x^2+ax+1=0$

This equation has a real root if and only if $\Delta=a^2-4\ge0$ or $a\ge2$ , since $a>0$ .

So, the minimum value of $1000a$ is $\boxed{2000}$ .