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$\frac{x^8-x^5}{x^7+x^6+x^5}$ $=$ $\frac{x^5(x^3-1)}{x^5(x^2+x+1}$ $=$ $\frac{(x-1)(x^2+x+1)}{(x^2+x+1)}$ $=$ $x-1$

If $x=2016$ then $x-1=2015$

Thus $\frac{2016^8-2016^5}{2016^7+2016^6+2016^5}$ $=$ $2015$

Taking $2016^5$ from both numerator and denominator:

$\dfrac{2016^{3(n-1)}-1}{2016^{2(n-1)}+2016^{(n-1)}+1}$

$=\dfrac{(\color{#D61F06}{2016^{(n-1)}})^3-1}{(\color{#D61F06}{2016^{(n-1)}})^2+\color{#D61F06}{2016^{(n-1)}} +1 }$

Substitute $\color{#D61F06}{2016^{(n-1)}}=\color{#D61F06}{t}$ and using the identity $a^3-1=(a-1)(a^2+a+1)$ or $\dfrac{a^3-1}{a^2+a+1}=a-1$ s.t expression simplifies to:

$=\large{2016^{(n-1)}-1}$

For $n=2$ , this is simply:

$\huge = \boxed{2015}$