If `f(x) = x^3 -3x + 1`, then the number of distinct real roots of the equation `f (f (x)) = 0` is

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`f'(x)=3x^2-3=0=>x=+-1`

Also `f''(x)=6x` so that `f''(1)>0` and `f''(-1)<0`.

Hence f is maximum at x=-1 and minimum at x=1. Hence f increase in `(-oo,-1)`, decreases in (-1,1) and again increasees in `(1,oo)`. Since `f(x)=0` has three zeros , say `alpha,beta,gamma` suppose `alpha<-10)`

Since f increases from `-oo` to 3 in `(-oo,-1)`, f assumes the values `alpha,beta,gamma` exactly once. Therefore `fof` has three zeroes in `(-oo,-1)`.Since f decreases from 3 to -1(-1,1), f assumes the values `beta ,gamma` only once in (-1,1). Lastly f increases from -1 to `oo` in `(1,oo)` , `fof` has two zeros in `(1,oo)`. Thus the total number of zeros of `fof` is 7.