Let $f(x)=x^{3}-12x$ be function such that the equation $|f(|x|)|=n(n\ \in\ N)$ has exactly $6$ distinct real roots then number of possible values of $n$ are :

Since the equation is $|x^3 - 12x| = n$, we require the horizontal line $y = n$ to intersect the graph of $y = |x^3 - 12x|$ at exactly six points. The cubic $x^3 - 12x$ has local extrema at $x = 2$ and $x = -2$, yielding extreme values of $-16$ and $16$, respectively. The absolute value graph will have local maxima at $16$ and a local minimum at $0$ (at $x = 0$), so $n$ must lie strictly between $0$ and $16$. Because $n$ is a natural number, it can take any integer value from $1$ to $15$, giving $15$ possible values.

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