Triple System of Equations

Note that we have

$\begin{aligned}\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma&=(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\alpha\gamma-\beta\gamma)\\&=(\alpha+\beta+\gamma)[(\alpha+\beta+\gamma)^2-3(\alpha\beta+\alpha\gamma+\beta\gamma)]\\&=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)\\\implies 29-\alpha\beta\gamma&=72-6(\alpha\beta+\alpha\gamma+\beta\gamma).\end{aligned}$

Additionally, expanding the third equation gives $1+\alpha+\beta+\gamma+\alpha\beta+\alpha\gamma+\beta\gamma+\alpha\beta\gamma=33,$ which implies $\alpha\beta+\alpha\gamma+\beta\gamma+\alpha\beta\gamma=26.$ Letting $\alpha\beta+\alpha\gamma+\beta\gamma=p$ and $\alpha\beta\gamma=q$ , we have the system of equations $\begin{cases}29-q&=72-6p,\\p+q&=26\end{cases}$ which after solving gives $(p,q)=(\frac{69}7,\frac{113}7)$ . Hence $\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\alpha\beta+\alpha\gamma+\beta\gamma}{\alpha\beta\gamma}=\dfrac{69/7}{113/7}=\dfrac{69}{113}$ and the requested answer is $69+113=\boxed{182}.$

Assume that $S_1=\alpha+\beta+\gamma, S_2=\alpha\beta+\beta\gamma+\gamma\alpha, S_3=\alpha\beta\gamma$ , and also $P_n=\alpha^n+\beta^n+\gamma^n$ , we have $S_1=P_1=6, P_3=87,$ and $S_2+S_3=26$ .

Using Newton's Identity : $P_n=S_1P_{n-1}-S_2P_{n-2}+S_3P_{n-3}$ , we have $\begin{aligned}P_2&=S_1^2-2S_2\\&=36-2S_2\end{aligned}$

Considering cubic, we have $\begin{aligned}P_3&=S_1P_2-S_2P_1+3S_3\\87&=216-12S_2-6S_2+3S_3\\\therefore 6S_2-S_3&=43\end{aligned}$ Now we have two equations about $S_2$ and $S_3$ , we'll get $S_2=\dfrac{69}{7}$ and $S_3=\dfrac{113}7$ after solving the system.

Then the question is ask for $\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\dfrac{S_2}{S_3}=\boxed{\dfrac{69}{113}}$ , the answer is $69+113=\boxed{182}$ .