Using Vieta's Formula? (2)

Relevant wiki: Vieta's Formula - Higher Degrees

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

$2017\alpha^{3}+2016\alpha^{2}+2015\alpha=-2016...(1)$

$2017\beta^{3}+2016\beta^{2}+2015\beta=-2016...(2)$

$2017\gamma^{3}+2016\gamma^{2}+2015\gamma=-2016...(3)$

LHS of $(1)-(2)-(3)$ gives the numerator while denominator is simply fourth root of $(\alpha\beta\gamma)^3(\alpha+\beta+\gamma)$ , but Vieta's formula give:

$\small{\color{#D61F06}{\alpha+\beta+\gamma=\alpha\beta\gamma=\dfrac{-2016}{2017}}}$ so that denominator is simply fourth root of $(\alpha\beta\gamma)^4$ which is $|\alpha\beta\gamma|$ . Therefore answer is:

$\large \dfrac{\cancel{2016}}{\frac{\cancel{2016}}{2017}}=\boxed{2017}$

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

$\alpha+\beta+\gamma = -\frac{2016}{2017}$

$\alpha\beta\gamma = (-1)^3\frac{2016}{2017} = -\frac{2016}{2017}$

$\dfrac{2017(\alpha^3-\beta^3-\gamma^3)+2016(\alpha^2-\beta^2-\gamma^2)+2015(\alpha-\beta-\gamma)}{\sqrt[4]{\alpha^4\beta^3\gamma^3+\alpha^3\beta^4\gamma^3+\alpha^3\beta^3\gamma^4}}$

$= \dfrac{-2016+2016+2016}{\sqrt[4]{(\alpha\beta\gamma)^3(\alpha+\beta+\gamma)}}$

$= \dfrac{2016}{\sqrt[4]{(\frac{2016}{2017})^4}}$

$= \dfrac{2016}{\frac{2016}{2017}}$

$= 2017$