Let's see what you've got

clearly sum of roots is zero.

$\alpha+\beta+\gamma=0$

$\alpha+\beta=-\gamma$ ...... $(1)$

$\alpha+\gamma=-\beta$ ......... $(2)$

$\beta+\gamma=-\alpha$ ......... $(3)$

Rearranging the given expression

$\frac{\alpha+\beta}{\gamma}+\frac{\alpha+\gamma}{\beta}+\frac{\beta+\gamma}{\alpha}$

From $(1),(2),(3)$

$\frac{-\gamma}{\gamma}+\frac{-\beta}{\beta}+\frac{-\alpha}{\alpha}$

=> $-1-1-1=\boxed{-3}$

My method resembles to that of tanishq but here's how I approached

For the convenience of my typing I am taking roots as a,b,c.

a+b+c=0

Summing the expression given we get

$\frac{x+y}{z}+\frac{y+x}{x}+\frac{z+x}{y}$

adding 3 and subtracting 3 we get

(x+y+z) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ -3

Since x+y+z=0 so value of the expression is $\boxed{-3}$

Let $\dfrac{\alpha}{\gamma}+\dfrac{\alpha}{\beta}+\dfrac{\gamma}{\beta}+\dfrac{\gamma}{\alpha}+\dfrac{\beta}{\alpha}+\dfrac{\beta}{\gamma} = S$

Multiplying $S$ by $\alpha\beta\gamma$ :

$\alpha\beta\gamma S = \alpha^2\beta+\alpha^2\gamma+\gamma^2\alpha+\gamma^2\beta+\beta^2\gamma+\beta^2\alpha$

$= \alpha\beta(\alpha+\beta)+\alpha\gamma(\alpha+\gamma)+\beta\gamma(\beta+\gamma)$

$=\alpha\beta(\alpha+\beta+\gamma-\gamma)+\alpha\gamma(\alpha+\gamma+\beta-\beta)+\beta\gamma(\beta+\gamma+\alpha-\alpha)$

By Vieta's, $\alpha\beta\gamma = -175$ and $\alpha+\beta+\gamma = 0$ , so:

$-175S = \alpha\beta(0-\gamma)+\alpha\gamma(0-\beta)+\beta\gamma(0-\alpha)$

$= \alpha\beta(-\gamma)+\alpha\gamma(-\beta)+\beta\gamma(-\alpha)$

$\implies -175S = -3\alpha\beta\gamma = 525$

$\implies S = \dfrac{525}{-175} = \boxed{-3}$

$Clearly\quad the\quad sum\quad of\quad roots\quad is\quad zero\\ \Rightarrow \alpha +\beta +\gamma =0\quad \quad \quad \quad \quad \quad \quad (1)\\ \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } =?\quad \quad \quad \quad (2)\\ (1)\times (2)\\ \left( \alpha +\beta +\gamma \right) \left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } \right) =1+\frac { \alpha }{ \beta } +\frac { \alpha }{ \gamma } +\frac { \beta }{ \alpha } +1+\frac { \beta }{ \gamma } +\frac { \gamma }{ \alpha } +\frac { \gamma }{ \beta } +1\\ \Rightarrow 0=\frac { \alpha }{ \beta } +\frac { \alpha }{ \gamma } +\frac { \beta }{ \alpha } +\frac { \beta }{ \gamma } +\frac { \gamma }{ \alpha } +\frac { \gamma }{ \beta } +3\\ \Rightarrow \frac { \alpha }{ \beta } +\frac { \alpha }{ \gamma } +\frac { \beta }{ \alpha } +\frac { \beta }{ \gamma } +\frac { \gamma }{ \alpha } +\frac { \gamma }{ \beta } =(-3)$