Using Vieta's Formula?

Relevant wiki: Vieta's Formula - Higher Degrees

Since $\alpha,\beta,\gamma$ are roots if the given equation, hence:

$2016\alpha^{3}+2015\alpha^{2}+2014\alpha=-2016...(1)$

$2016\beta^{3}+2015\beta^{2}+2014\beta=-2016...(2)$

$2016\gamma^{3}+2015\gamma^{2}+2014\gamma=-2016...(3)$

LHS of $(1)+(2)+(3)$ gives the numerator while LHS of $(1)-(2)-(3)$ gives the denominator , hence answer is:

$\large \dfrac{-(2016+2016+2016)}{-(2016-2016-2016)}=\boxed{-3}$

Relevant wiki: Vieta's Formula - Higher Degrees

Since $2016x^{3}+2015x^{2}+2014x+2016 = 0$ $\implies 2016x^{3}+2015x^{2}+2014x = -2016$ . And since $\alpha$ , $\beta$ and $\gamma$ are the roots of the equation, then we have:

$\begin{cases} 2016 \alpha^{3}+2015\alpha^{2}+2014\alpha = -2016 \\ 2016 \beta^{3}+2015\beta^{2}+2014\beta = -2016 \\ 2016 \gamma^{3}+2015\gamma^{2}+2014\gamma = -2016 \end{cases}$

Therefore,

$\dfrac{2016(\alpha^3+\beta^3+\gamma^3)+2015(\alpha^2+\beta^2+\gamma^2)+2014(\alpha+\beta+\gamma)}{2016(\alpha^3-\beta^3-\gamma^3)+2015(\alpha^2-\beta^2-\gamma^2)+2014(\alpha-\beta-\gamma)} = \frac {-2016-2016-2016}{-2016+2016+2016} = \boxed{-3}$

$\begin{cases} 2016\alpha^{3}+2015\alpha^{2}+2014\alpha+2016=0 \\ 2016\beta^{3}+2015\beta^{2}+2014\beta+2016=0 \\ 2016\gamma^{3}+2015\gamma^{2}+2014\gamma+2016=0 \end{cases}$

$\begin{cases} 2016\alpha^{3}+2015\alpha^{2}+2014\alpha=-2016 \\ 2016\beta^{3}+2015\beta^{2}+2014\beta=-2016 \\ 2016\gamma^{3}+2015\gamma^{2}+2014\gamma=-2016 \end{cases}$

$\dfrac{2016(\alpha^3+\beta^3+\gamma^3)+2015(\alpha^2+\beta^2+\gamma^2)+2014(\alpha+\beta+\gamma)}{2016(\alpha^3-\beta^3-\gamma^3)+2015(\alpha^2-\beta^2-\gamma^2)+2014(\alpha-\beta-\gamma)}$

$=\dfrac{(-2016)\times3}{(-2016)-(-2016)\times2}$

$= - \dfrac{6048}{2016}$

$=-3$