Evaluate $\displaystyle \left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$.

Evaluate $\displaystyle \left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$.

Let $x=2015$.

Therefore we see that we have:

$\displaystyle \begin{align*}\left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor&=\left\lfloor{\frac{(x-1)^3}{x(x+1)}+\frac{(x+1)^3}{(x-1)(x)}}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{(x-1)^3}{x+1}+\frac{(x+1)^3}{x-1}\right)}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{(x-1)^4+(x+1)^4}{(x-1)(x+1)}\right)}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{((x-1)^2)^2+((x+1)^2)^2}{x^2-1}\right)}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{((x-1)^2-(x+1)^2)^2+2(x-1)^2(x+1)^2}{x^2-1}\right)}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{((2x)(-2))^2+2(x^2-1)^2}{x^2-1}\right)}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{((x-1+x+1)(x-1-x-1))^2+2(x^2-1)^2}{x^2-1}\right)}\right\rfloor\\&=\left\lfloor{\frac{1}{x}\left(\frac{16x^2+2(x^2-1)^2}{x^2-1}\right)}\right\rfloor\\&=\left\lfloor{\frac{16x^2}{x(x^2-1)}+\frac{2(x^2-1)^2}{x(x^2-1)}}\right\rfloor\\&=\left\lfloor{\frac{16x}{x^2-1}+\frac{2(x^2-1)}{x}}\right\rfloor\\&=\left\lfloor{\frac{16x}{x^2-1}+\frac{2x^2}{x}-\frac{2}{x}}\right\rfloor\\&=\left\lfloor{2x+\frac{16x}{x^2-1}-\frac{2}{x}}\right\rfloor\\&=2x\\&=2(2015)\\&=4030\end{align*}$

We can omit and ignore the difference of $\frac{16x}{x^2-1}-\frac{2}{x}$ in our last step since when $x=2015$, their difference must be a decimal.