It needs some construction

A complex number $a+ib$ for which $|a:b|=1:√3\space or \space √3:1$ can always be put in terms of $\omega$ (the second cube root of unity).

Now $(i+√3)=-2i\omega$ $(i-√3)=-2i\omega^2$ (Try proving them)

So that our required expression becomes, $2^{100}\omega+ 2^{100}\omega^2+ 2^{100}$ $=2^{100}(\omega+\omega^2)+ 2^{100}$ $= 2^{100}(-1)+ 2^{100}=\boxed 0$

We can use the Euler's identity: $e^{i\theta} = \cos {\theta} + i\sin{\theta}$

$(i+\sqrt{3})^{100} + (i-\sqrt{3})^{100} +200^{100}$

$=2^{100}(\frac {\sqrt{3}}{2}+i\frac {1}{2} )^{100} + 2^{100}(\frac {-\sqrt{3}}{2}+i\frac {1}{2} )^{100} +200^{100}$

$=2^{100} \left[ (\cos{\frac {\pi}{6} }+i \sin {\frac {\pi}{6} })^{100} + (\cos{\frac {5\pi}{6} }+i \sin {\frac {5\pi}{6} } )^{100} + 1 \right]$

$=2^{100} \left[ (e^{i \frac {\pi}{6}}) ^{100} + (e^{i\frac {5\pi}{6}}) ^{100} + 1 \right] = 2^{100} (e^{i \frac {100\pi}{6}} + e^{i\frac {500\pi}{6}} + 1 )$

$=2^{100} (e^{i16 \frac {2\pi}{3}} + e^{i83\frac {\pi}{3}} + 1 ) =2^{100} (e^{i \frac {2\pi}{3}} + e^{i\frac {4\pi}{3}} + 1 )$

$=2^{100} (\cos{\frac {2\pi}{3} }+i \sin {\frac {2\pi}{3} } + \cos{\frac {4\pi}{3} }+i \sin {\frac {4\pi}{3} } + 1 )$

$=2^{100}( -\frac {1}{2} + i\frac {\sqrt{3}}{2} - \frac {1}{2} - i\frac {\sqrt{3}}{2}+1 ) = \boxed {0}$

Using De Moivre's Thm: (3^0.5 + i)^100 + (3^0.5 - i)^100 +2^100 = 2^100(cis(100pi/6)) - 2^100(cis(-100pi/6)) + 2^100 = 2^100(-0.5 + (3^0.5)(i/2)) - 2^100(0.5 - (3^0.5)(i/2)) + 2^100 = 2^100 - 2^100 = 0

I used de Moivre after rewriting in polar form. Then it was straightforward, using the periodicity of trigonometric functions and some factorization, but not as elegant as the previous solution.

(-2iw)^100+(-2iw^2)^2+2^100=(2w)^100+(2w^2)^100+2^100=2^100(w+w^2+1)=2^100(0)=0