PP!!

$\alpha=a , \beta=b$

$x^2 - 3x + 1=0$

$a^2 - 3a + 1 =0$

$b^2 - 3b + 1 =0$

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

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

$(a^{2016} + b^{2016}) + (a^{2014} + b^{2014}) = 3(a^{2015} + b^{2015})$

$Given ~ expression ~ = 3$

Let ${ P }_{ n }\quad =\quad \alpha ^{ n }+{ \beta }^{ n }\\ { S }_{ 1 }\quad =\quad \alpha +\beta \\ { S }_{ 2 }\quad =\quad \alpha \beta$

Now, Using Newton's Sums: ${ P }_{ n }\quad ={ \quad P }_{ n-1 }{ S }_{ 1 }\quad -\quad { P }_{ n-2 }{ S }_{ 2 },\quad \forall \quad n\ge 3$

In our case, ${ S }_{ 1 }\quad =\quad 3,\quad { S }_{ 2 }\quad =\quad 1$

Now, in the given fraction, $\frac { { \alpha }^{ 2014 }+{ \beta }^{ 2014 }+{ \alpha }^{ 2016 }+\beta ^{ 2016 } }{ { \alpha }^{ 2015 }+{ \beta }^{ 2015 } } \\ \\ =\quad \frac { { P }_{ 2014 }+{ P }_{ 2016 } }{ { P }_{ 2015 } } \\ \\ =\quad \frac { { P }_{ 2014 }+{ P }_{ 2015 }{ S }_{ 1 }-P_{ 2014 }{ S }_{ 2 } }{ { P }_{ 2015 } } \\ \\ =\quad \frac { { P }_{ 2014 }+{ P }_{ 2015 }(3)-P_{ 2014 }(1) }{ { P }_{ 2015 } } \\ \\ =\quad \frac { { P }_{ 2015 }(3) }{ { P }_{ 2015 } } \\ \\ =\quad 3$

CHHERS!:)