The value of (α^2/(βγ)+(β^2/(γα)+(γ^2/(αβ) is

Since $\alpha\beta\gamma=1$ (product of roots of the monic cubic is $-(-1)=1$), each term simplifies as $\dfrac{\alpha^2}{\beta\gamma}=\dfrac{\alpha^3}{\alpha\beta\gamma}=\alpha^3$. So the required sum equals $\alpha^3+\beta^3+\gamma^3$. With $e_1=\sum\alpha=4$, $e_2=\sum\alpha\beta=5$, $e_3=\alpha\beta\gamma=1$, the identity $\sum\alpha^3=e_1^3-3e_1e_2+3e_3$ gives $64-60+3=7$. Forgetting the $3e_3$ term gives the trap value $4$, while $19$ is the value $4^3-4\cdot4^2+5\cdot4-1$ obtained by wrongly evaluating the cubic at $x=e_1$. A numerical check of the three roots confirms the sum is $7$. Hence the answer is $7$. This is a typical Advanced-style reduction: recognise the hidden symmetric function, convert it to a power sum, and then use Newton's identities rather than finding the roots.