How do i draw the bode plot for H(s) = 100000 ∙ (s + 10) ^3 /( s ∙ (s + 100) ∙ (s + 1000j) ∙ (s − 1000j))
Substitute \$s = j\omega\$, and plot the value for \$|H(j\omega)|\$. 
For the \$ j\$ question, \$ j^2 = -1\$. 

Here you can find a good lesson and some quick resource on how to plot the bode plot of signals. 

So, we start with,
$$ H(s) =  \frac{100000(s + 10) ^3 }{s(s + 100)(s + 1000j)(s − 1000j)}, $$
and replace \$ s = j\omega\$, resulting in,
$$ H(j\omega) =  \frac{100000(j\omega + 10) ^3 }{j\omega(j\omega + 100)(j\omega + 1000j)(j\omega − 1000j)}, $$

that can be further simplified to get
$$ H(j\omega) =  \frac{100000 j(j\omega + 10) ^3 }{\omega(j\omega + 100)(\omega + 1000)(\omega − 1000)}. $$

The magnitude of the signal is:
$$ \left| H(j\omega)\right| = \frac{100000 (\omega^2+100)\sqrt{\omega^2+100}}{\omega \left(\omega + 1000\right) \sqrt{\omega^{2} + 10000} \left|{\omega - 1000}\right|}$$

Now, you either put that in to some software or plot it manually, I used octave to get this get this:

f = @(w) (100000*(w.^2+100).*sqrt(w.^2+100))./(w.*(w+1000).*(w.^2+10000).*abs(w-1000))
t = 10.^(0.001:0.001:5);
loglog(t,f(t))


Resulting in,


Which resembles a lot the approximation using the asymptotes, the greatest differences are near the zeros and poles, where it differs the most. Also, this plot does not reproduce completely the peak at \$ \omega = 1000 \text{ rad/s}\$. But as you change the resolution of the plot (the values used for \$ t\$) it becomes higher and higher (infinity gain).