First, this is the basic function of variable replacement. It can be understood that given the value of x, finding f(x) means replacing variables with constants. Since we can use x- 1 instead of x, we can also use x instead of x- 1.

The second problem is that knowing that f(x) is a linear function can make f (x) = ax+b.

So we can get f (f (x)) = f (ax+b) = a (ax+b)+b = a2x+ab+b.

f(f(f(x)))=f(a^2x+ab+b)=a(a^2x+ab+b)+b=a^3x+a^2b+ab+b=27x+36

This formula is an identity, so the coefficients and constants of x on both sides should be equal, so we have:

a^3=27

a^2b+ab+b=36 9b+3b+b=36

Solution: a=3 b=36/ 13

Maybe I miscalculated, but I gave you the idea.