Bashing roots

Since $a$ is a root we have : $a^2-3a+1=0$

Let's see what we get after multiplying both sides by $a^{2014}$ . Indeed we get the following expression:

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

Similarly we get :

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

Adding both the above equations , we get :

$a^{2014}+b^{2014}+a^{2016} + b^{2016}-3a^{2015}-3b^{2015}=0 \\ \Rightarrow a^{2014}+b^{2014}+a^{2016} + b^{2016}=3(a^{2015}+b^{2015}) \\ \Rightarrow \dfrac{a^{2014} + b^{2014} + a^{2016} + b^{2016}}{a^{2015} + b^{2015}} = \huge\boxed{3}$

If $a$ and $b$ are the roots of the given quadratic equation

$\implies a^2-3a+1=0$ and $b^2-3b+1=0$

$\implies a^2+1=3a$ and $b^2+1=3b$ ............................... (1)

Now on rearranging the terms in the numerator the given expression is:

$\dfrac{a^{2016}+a^{2014}+b^{2016}+b^{2014}}{a^{2015}+b^{2015}}$

$=\dfrac{a^{2014}(a^2+1)+b^{2014}(b^2+1)}{a^{2015}+b^{2015}}$

$=\dfrac{a^{2014}.3a+b^{2014}.3b}{a^{2015}+b^{2015}}$ Using (1)

$=\dfrac{3(a^{2015}+b^{2015})}{(a^{2015}+b^{2015})}$

$=\boxed{3}$

Let $x_n=a^n+b^n$ . According to the theory of Binet Forms, we have the recursive equation $x_n=3x_{n-1}-x_{n-2}$ , or, $\frac{x_{n-2}+x_{n}}{x_{n-1}}=3$ . Applying this equation to $n=2016$ , we find the answer, $\boxed{3}$ .

$a+b=3~~and~~~~~ ab=1.\\(a^{2015} + b^{2015})*(a+b)=a^{2016} + b^{2016}+ab*(a^{2014} + b^{2014})\\\implies~a^{2015} + b^{2015}=\dfrac{a^{2016} + b^{2016}+1*(a^{2014} + b^{2014})}{3} \\ \therefore~f(a,b)=3$