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In exponential notation, 1 + i*sqrt(3) = sqrt(3)*e^(i*pi/3).  Then

(1 + i*sqrt(3))^6 = (sqrt(3)*e^(i*pi/3))^6 = 27.

$\displaystyle 1 + i\sqrt{3} = 2(1/2 + i\sqrt{3}/2)=2(\cos(\pi/3)+i\sin(\pi/3)). \\ (1+i\sqrt{3})^{6}=\{2(\cos(\pi/3)+i\sin(\pi/3))\}^{6}=2^{6}(\cos(2\pi)+i\sin(2\pi)) \\ =2^{6}=64.$