Vieta's is ok

Putting $\displaystyle x=\frac{1}{t}$ , we get :

$\displaystyle \Rightarrow t^6 - t^5 + 3t^4 + 43t^2 + 4t + 1$

The roots of this equation are $\frac{1}{a_n}$ .

Applying Vieta's formula :

$\displaystyle \Rightarrow \displaystyle \sum \frac{1}{a_n} = \boxed{1}$

$\large \displaystyle \sum_{cyc}\dfrac{1}{a_1}= \dfrac{\displaystyle \sum_{cyc}a_1a_2a_3a_4a_5}{\displaystyle \prod_{cyc} a_1}$ $\large ~~~~~~~~~~~~~~~~~~=\dfrac{1}{1}....(\small{using~Vieta's)}$ $\huge=\boxed{\color{#007fff}{1}}$

Since the constant term is 1, the polynomial can be factored as $\Big(1-\frac{x}{a_1}\Big)\Big(1-\frac{x}{a_2}\Big)...\Big(1-\frac{x}{a_6}\Big)$ . Then notice through this factorization (think Vieta's formulas) that the linear term is the negative of the sum of the reciprocal of the roots. Therefore, the desired quantity is the negative of the coefficient of the linear term, i.e., 1.