Powers of 2014 , 2015 and 2016

$\alpha^2 - 3\alpha + 1 =0$

Multiplying by $\alpha^{2014}$

$\alpha^{2016} - 3\alpha^{2015} + \alpha^{2014} = 0$

Similarly

$\beta^{2016} - 3\beta^{2015} + \beta^{2014} = 0$

Adding,

$\alpha^{2014} + \beta^{2014} + \alpha^{2016} + \beta^{2016} = 3( \alpha^{2015} + \beta^{2015})$

$\dfrac{\alpha^{2014} + \beta^{2014} + \alpha^{2016} + \beta^{2016} }{\alpha^{2015} + \beta^{2015}} = 3$

The expreesion to be found out cna be written as

$\dfrac{\alpha^{2014}(\alpha^{2}+1)+\beta^{2014}(\beta^{2}+1)}{\alpha^{2015}+\beta^{2015}}$

Since $\alpha \ \& \ \beta$ are the roots of $x^{2}-3x+1=0$

So $\alpha^{2}-3\alpha+1=0$ and $\beta^{2}-3\beta+1=0$

By this $\alpha^{2}+1=3\alpha$ and $\beta^{2}+1=3\beta$

Substitute these values in the topmost expression to get

$\dfrac{\alpha^{2014}(3\alpha)+\beta^{2014}(3\beta)}{\alpha^{2015}+\beta^{2015}}$

$\Rightarrow \dfrac{ 3(\alpha^{2015}+\beta^{2015})}{\alpha^{2015}+\beta^{2015}}$

So the asnwer is $\huge 3$

the roots of the polynomial are $\varphi^2,\varphi^{-2}$ by identites of the golden ratios, $\varphi^n+(-\varphi)^{-n}=L_{n}$ such that $L_{n} =\begin{cases} 2, n=0\\ 1,n=1\\ L_{n-1}+L_{n-2},n\geq 2 \end{cases}$ hence, $\dfrac{\varphi^{4028}+\varphi^{-4028}+\varphi^{4032}+\varphi^{-4032}}{\varphi^{4030}+\varphi^{-4030}}=\dfrac{L_{4028}+L_{4032}}{L_{4030}}$ rewrite $\dfrac{L_{4028}+L_{4032}}{L_{4030}}=\dfrac{(L_{4030}-L_{4029})+(L_{4030}+L_{4031})}{L_{4030}}=\dfrac{2L_{4030}+L_{4031}-L_{4029}}{L_{4030}}=\dfrac{3L_{4030}}{L_{4030}}=\boxed{3}$

I just thought: alpha is between 0 and 1, beta is between 2 and 3, the numerator and the denominator are mainly defined by giant powers of beta so at least the answer is beta, but it's a tiny bit more because of a lower power in sum above, and it couldn't be more than beta squared; so the plausible rough answers were 3, 4, and 5; 5 seemed too close to beta squared, four seemed too much, 3 made sense.

Using the Newton's Sums Method, we see that for n >= 3, S(n) = 3S(n-1)-S(n-2), where S(n) = a^n + b^n. Thus, S(2016) = 3S(2015)-S(2014), and the fraction becomes (S(2014)+S(2016))/S(2015) = 3S(2015)/S(2015) = 3.

$\frac { { \alpha }^{ 2014 }+{ \beta }^{ 2014 }+{ \alpha }^{ 2016 }+{ \beta }^{ 2016 } }{ { \alpha }^{ 2015 }+{ \beta }^{ 2015 } } \quad =\quad \frac { { \alpha }^{ 2015 }(\alpha +\cfrac { 1 }{ \alpha } )+{ \beta }^{ 2015 }(\beta +\cfrac { 1 }{ \beta } ) }{ { \alpha }^{ 2015 }+{ \beta }^{ 2015 } } .\quad \\ \Rightarrow \quad \frac { 3({ \alpha }^{ 2015 }+{ \beta }^{ 2015 }) }{ ({ \alpha }^{ 2015 }+{ \beta }^{ 2015 }) } \quad =\quad 3.\quad \quad$