[b]5.[/b] Find the continuous solutions of the functional equation $f(xyz)= f(x)+f(y)+f(z)$ in the following cases:

(a) $x,y,z$ are arbitrary non-zero real numbers;
(b) $a<x,y,z<b (1<a^{3}<b)$. 

[b](R. 13)[/b]
1. **Part (a):**

   The given functional equation is:
   \[
   f(xyz) = f(x) + f(y) + f(z)
   \]
   for arbitrary non-zero real numbers \(x, y, z\).

   To solve this, we can take the logarithmic map. Let \(x = e^u\), \(y = e^v\), and \(z = e^w\). Then, \(xyz = e^{u+v+w}\). Define a new function \(F\) such that \(F(u) = f(e^u)\). The functional equation becomes:
   \[
   f(e^{u+v+w}) = f(e^u) + f(e^v) + f(e^w)
   \]
   which translates to:
   \[
   F(u+v+w) = F(u) + F(v) + F(w)
   \]
   This is a form of Cauchy's functional equation. The continuous solutions to Cauchy's functional equation are of the form \(F(x) = \lambda x\) for some constant \(\lambda\). Therefore, we have:
   \[
   f(e^u) = \lambda u
   \]
   or equivalently:
   \[
   f(x) = \lambda \ln x
   \]
   for some constant \(\lambda\).

2. **Part (b):**

   Here, \(x, y, z\) are in the interval \((a, b)\) with \(1 < a^3 < b\). We interpret the function \(f: (a, b) \rightarrow \mathbb{R}\) which satisfies \(f(xyz) = f(x) + f(y) + f(z)\) whenever \(x, y, z, xyz \in (a, b)\).

   Define \(A = \ln a\) and \(B = \ln b\). Let \(F(x) = f(e^x)\) for \(A < x < B\). The functional equation becomes:
   \[
   F(x + y + z) = F(x) + F(y) + F(z)
   \]
   for \(A < x, y, z < B\), with the condition \(0 < 3A < B\).

   For any \(0 < \epsilon < B - 3A\), define \(D_\epsilon: (A, B - 2A - \epsilon) \rightarrow \mathbb{R}\) by:
   \[
   D_\epsilon(x) = F(x + \epsilon) - F(x)
   \]

   For any \(x, y > A\) such that \(x + y < B - A - \epsilon\), there exists \(z > A\) such that \(x + y + z + \epsilon < B\). Consequently, we have:
   \[
   F(x + \epsilon) + F(y) + F(z) = F(x + y + z + \epsilon) = F(x) + F(y + \epsilon) + F(z)
   \]
   which gives:
   \[
   D_\epsilon(x) = D_\epsilon(y)
   \]

   This implies that the function \(D_\epsilon(x)\) is constant for \(x \in (A, B - 2A - \epsilon)\). Denote this constant by \(d_\epsilon\).

   For \(\epsilon, \delta > 0\) such that \(\epsilon + \delta < B - 3A\), we have:
   \[
   d_{\epsilon + \delta} = d_\epsilon + d_\delta
   \]
   Since \(F\) is continuous, the function \(\epsilon \mapsto d_\epsilon\) is also continuous. The standard arguments of Cauchy's functional equation then show that there exists \(\lambda \in \mathbb{R}\) such that \(d_\epsilon = \lambda \epsilon\) for all \(0 < \epsilon < B - 3A\).

   Now, define \(G(x) = F(x) - \lambda x\). Then \(G\) is continuous and satisfies:
   \[
   G(x + \epsilon) - G(x) = 0
   \]
   for all \(\epsilon, x\) with \(0 < \epsilon < B - 3A\) and \(A < x < B - 2A - \epsilon\). This means that \(G(x)\) is constant for \(x \in (A, B - 2A)\). Denote this constant by \(g\).

   Since we still have \(G(x + y + z) = G(x) + G(y) + G(z)\), we have \(G(x) = 3g\) for \(3A < x < B\). Thus, there are two cases:

   1. If \(B - 2A \leq 3A\), i.e., \(B \leq 5A\), then \(g\) can be any real number, and we will have:
      \[
      F(x) = \begin{cases}
      \lambda x + g, & \text{if } x < B - 2A; \\
      \text{arbitrary}, & \text{if } B - 2A \leq x \leq 3A; \\
      \lambda x + 3g, & \text{if } 3A < x < B.
      \end{cases}
      \]
      It is easy to verify that this function satisfies the original condition.

   2. If \(B - 2A > 3A\), i.e., \(B > 5A\), then on the interval \((3A, B - 2A)\), we must have \(G(x) = g\) and \(G(x) = 3g\), which forces \(g = 0\). Therefore, we have \(F(x) = \lambda x\) in this case, which is clearly a valid solution.

The final answer is \( \boxed{ f(x) = \lambda \ln x } \) for part (a) and \(F(x) = \lambda x\) for part (b) with the specified conditions.