find $\operatorname{Var}(\hat\epsilon_i)=\sigma^2({{n-1}\over n}-{{(x_i-\bar x)^2}\over{\sum_{i=1}^n}(x_i-\bar x)^2})$

Question

I'm studying question 16.19 in the book Probability Essentials. The question is to show that: $\operatorname{Var}(\hat\epsilon_i)=\sigma^2({{n-1}\over n}-{{(x_i-\bar x)^2}\over{\sum_{i=1}^n}(x_i-\bar x)^2})$.
where $\hat\epsilon_i=Y_i-A-Bx_i$.
{$\epsilon_i$} is i.i.d. $N(0,\sigma^2)$.
$A,B$ is the estimator of the for $\alpha$ and $\beta$ in linear regression $Y_i=\alpha+\beta x_i+\epsilon_i$. \begin{align} \operatorname{Var}(A)&={\sigma^2\sum x_i^2\over n\sum (x_i-\bar x)^2}. \\ \operatorname{Var}(B)&={\sigma^2\over n\sum (x_i-\bar x)^2}. \\ \operatorname{Cov}(A,B)&={-\sigma^2\bar x\over n\sum (x_i-\bar x)^2} \end{align} My attempt: \begin{align} \operatorname{Var}(\hat\epsilon_i)&=\operatorname{Var}(Y_i-A-Bx_i)\\ &=\operatorname{Var}(Y_i)+\operatorname{Var}(A+Bx_i)\\ &=\sigma^2+\operatorname{Var}(A)+x_i^2\operatorname{Var}(B)+2x_i\operatorname{Cov}(X,Y) \end{align} After the substitution, I still couldn't get the answer. I guess $Y_i$ is not independent with A and B? But what's the correct way to do it?