Let $$f(x) = \begin{cases} x - 1, & x \text{ is even} \\ 2x, & x \text{ is odd} \end{cases}, \ x \in \mathbb{N}$$. If for some $$a \in \mathbb{N}$$, $$f(f(f(a))) = 21$$, then $$\lim_{x \rightarrow a^{-}} \left\{ \frac{\vert{}x\vert{}^3}{a} - \left[\frac{x}{a}\right] \right\}$$ where $$[t]$$ denotes the greatest integer less than or equal to $$t$$, is equal to:

$$f(x) = \begin{cases} x - 1; & x = \text{even} \\ 2x; & x = \text{odd} \end{cases}$$

$$f(f(f(a))) = 21$$

Case-1: If $$a = \text{even}$$

$$f(a) = a - 1 = \text{odd}$$

$$f(f(a)) = 2(a - 1) = \text{even}$$

$$f(f(f(a))) = 2a - 3 = 21 \Rightarrow a = 12$$

Case-2: If $$a = \text{odd}$$

$$f(a) = 2a = \text{even}$$

$$f(f(a)) = 2a - 1 = \text{odd}$$

$$f(f(f(a))) = 4a - 2 = 21 \text{ (Not possible)}$$

Hence $$a = 12$$

Now, $$\lim_{x \rightarrow 12^{-}} \left( \frac{\vert{}x\vert{}^3}{12} - \left[\frac{x}{12}\right] \right)$$

$$= \lim_{x \rightarrow 12^{-}} \frac{\vert{}x\vert{}^3}{12} - \lim_{x \rightarrow 12^{-}} \left[\frac{x}{12}\right]$$

$$= 144 - 0$$ $$= 144$$