Given $x^2-2√3 x+4=0$

Solving quadratic equation we get

$x=\dfrac {-(-2√3) ± \sqrt{(2√3)^2-4(4)}}{2(1) } \\ ∴ = \dfrac {(2√3±\sqrt{12-16}}2 \\ = \dfrac {2√3 ± 2i}2 \\ =2(\dfrac {√3}2 ± \dfrac i2) \\ =2(\cos \dfrac π6 ± i\sin \dfrac π6) \\ ∴α = 2(\cos\dfrac π6 + i\sin \dfrac π6) \\ β= 2(\cos\dfrac π6- i \sin\dfrac π6) \\ ∴α^3=[2(\cos\dfrac π6+i \sin \dfrac π6)]^3 \\ =2^3 [\cos\dfrac π2+i\sin\dfrac π2] \\ \text { Similarly } β^3=2^3 [\cos\dfrac π2-i\sin \dfrac π2] \\ ∴α^3+β^3 = 2^3 [\cos \dfrac π2+i\sin\dfrac π2 + \cos \dfrac π2-i\sin\dfrac π2] \\ =2^3 [0+0] \\ ∴α^3+β^3=0 $