Find all rational zeros of $f$. Then (if necessary) use the depressed equation to find all roots of the equation $f(x)=0$. \[ f(x)=x^{3}-18 x-8 \] Select the correct choice below and fill in the answer box within your choice. A. The set of all zeros of the given function is \{\} . (Simplify your answer. Type an exact answer, using radicals as needed. Use a comma to separate answer as needed.) B. There are no real zeros.

Step 1 ：Given the function \(f(x)=x^{3}-18 x-8\), we are to find all the zeros of the function.

Step 2 ：According to the rational root theorem, any rational root, \(\frac{p}{q}\) (where p and q are relatively prime), of a polynomial equation \(a_nx^n + a_{n-1}x^{n-1} + ... + a_2x^2 + a_1x + a_0 = 0\) must be such that p is a factor of \(a_0\) and q is a factor of \(a_n\). In this case, \(a_0 = -8\) and \(a_n = 1\).

Step 3 ：The possible rational roots are the factors of -8 divided by the factors of 1, which are \([-8, -4, -2, -1, 1, 2, 4, 8]\).

Step 4 ：By substituting these possible roots into the function, we find that -4 is a rational root of the equation.

Step 5 ：We can then use synthetic division to find the depressed equation, which is \(x^2 - 4x - 2\).

Step 6 ：Solving this depressed equation gives us the remaining roots, which are \(2 - \sqrt{6}\) and \(2 + \sqrt{6}\).

Step 7 ：Thus, the set of all zeros of the given function is \(\boxed{\{-4, 2 - \sqrt{6}, 2 + \sqrt{6}\}}\).