$x^3+x^2+x+1=0$
$\alpha, \beta, \gamma$ are roots
$\alpha+\beta+\gamma=-1 ; \alpha \beta+\beta \gamma+\gamma \alpha=1 ; \alpha \beta \gamma=-1$
$\begin{aligned} & \text { (i) } \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\alpha \beta}{}+\frac{\beta \gamma+r \alpha}{\alpha \beta \gamma} \\ & =\frac{1}{-1}=-1 \Rightarrow \text { (i) } \rightarrow A\end{aligned}$
$\begin{aligned} & \text { (ii) } \alpha^3+\beta^3+\gamma^3 \\ & =(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)+3 \alpha \beta \gamma \\ & =(\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^2-3(\alpha \beta+\beta \gamma+\gamma \alpha)\right]+3 \alpha \beta \gamma\end{aligned}$
$=-1\left[(-1)^2-3(1)\right]+3(-1)=2-3=-1$
(ii) $\rightarrow A$
only possible option is
(i) $\rightarrow A$, (ii) $\rightarrow A$, (iii) $\rightarrow D$, (iv) $\rightarrow B$