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$\begin{aligned} X & = \frac {2015\times 4033^3+6045}{2016^3-1} + \frac {2017\times 4031^3+6051}{2016^3+1} & \small {\color{#3D99F6}\text{Let }a=2016} \\ & = \frac {(a-1)(2a+1)^2+3(a-1)}{a^3-1} + \frac {(a+1)(2a-1)^2+3(a+1)}{a^3+1} \\ & = \frac {(2a+1)^2+3}{a^2+a+1} + \frac {(2a-1)^2+3}{a^2-a+1} \\ & = \frac {4a^2+4a+4}{a^2+a+1} + \frac {4a^2-4a+4}{a^2-a+1} \\ & = 4+4 \\ & = \boxed{8} \end{aligned}$

$=\dfrac{2015\times{4033}^2+2015\times 3}{{2016}^3-1}+\dfrac{2017\times {4031}^2+2017\times 3}{{2016}^3+1}$

We now use ${\color{#D61F06}(a^3+b^3)=(a+b)(a^2-ab+b^2)}$ and ${\color{#3D99F6}(a^3-b^3)=(a-b)(a^2+ab+b^2)}$ to factorize the denominators.

$=\dfrac{\cancel{2015}\times({4033}^2+3)}{\cancel{(2016-1)}({2016}^2+2016+1)}+\dfrac{\cancel{2017}\times({4031}^2+3)}{\cancel{(2016-1)}({2016}^2-2016+1)}$

$=\dfrac{{(2016\times2+1)}^2+3}{{2016}^2+2016+1}+\dfrac{{(2016\times2-1)}^2+3}{{2016}^2-2016+1}$

$=\dfrac{4\times {2016}^2+4\times 2016+4}{{2016}^2+2016+1}+\dfrac{4\times {2016}^2-4\times 2016+4}{{2016}^2-2016+1}$

$=\dfrac{4\times(\cancel{{2016}^2+2016+1})}{(\cancel{{2016}^2+2016+1})}+\dfrac{4\times(\cancel{{2016}^2-2016+1})}{(\cancel{{2016}^2-2016+1})}$

$=4+4=\boxed{8}.$