Given equation $x^2-2 \sqrt{3} x+4=0$ with $\alpha \& \beta$ roots of the equation.
$\begin{aligned}
& \alpha+\beta=\frac{-b}{a}=2 \sqrt{3} \\
& \alpha \beta=\frac{c}{a}=4
\end{aligned}$
Now, $\left(\alpha^6+\beta^6\right)=\left(\alpha^2\right)^3+\left(\beta^2\right)^3$
$\begin{aligned}
& =\left(\alpha^2+\beta^2\right)\left[\alpha^4+\beta^4-\alpha^2 \beta^2\right] \\
& =\left(\alpha^2+\beta^2\right)\left[\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2-\alpha^2 \beta^2\right] \\
& =\left(\alpha^2+\beta^2\right)\left[\left(\alpha^2+\beta^2\right)^2-3 \alpha^2 \beta^2\right]
\end{aligned}$
From (i), take square both sides,
$\begin{aligned}
& \alpha^2+\beta^2+2 \alpha \beta=12 \\
& \alpha^2+\beta^2=12-2 \times 4=12-8=4
\end{aligned}$
Put the values in eq. (iii),
$\begin{aligned}
& \alpha^6+\beta^6=(4)\left[(4)^2-3(4)^2\right] \\
& =4[16-48] \\
& =4 \times(-32)=-128
\end{aligned}$
So, option (d) is correct.