Let the cubic polynomial be $$p(x) = a x^3 + b x^2 + c x + d$$ where \(a, b, c, d\) are real constants.

Given:
Local maximum at \(x = 1 \Rightarrow p'(1) = 0,\ p(1) = 8\)
Local minimum at \(x = 2 \Rightarrow p'(2) = 0,\ p(2) = 4\)

Derivative:
$$p'(x) = 3a x^2 + 2b x + c$$ Apply the stationary point conditions:
\[ \begin{cases} 3a(1)^2 + 2b(1) + c = 0 \\[4pt] 3a(2)^2 + 2b(2) + c = 0 \end{cases} \] Simplify: \[ \begin{cases} 3a + 2b + c = 0 \\[4pt] 12a + 4b + c = 0 \end{cases} \] Subtracting gives: \[ 9a + 2b = 0 \Rightarrow b = -\frac{9a}{2}. \] Substitute into \(3a + 2b + c = 0\): \[ 3a + 2\left(-\frac{9a}{2}\right) + c = 0 \Rightarrow 3a - 9a + c = 0 \Rightarrow c = 6a. \]
Using the value conditions:
\[ \begin{cases} p(1) = a + b + c + d = 8 \\[4pt] p(2) = 8a + 4b + 2c + d = 4 \end{cases} \] Substitute \(b = -\frac{9a}{2},\ c = 6a\):
\[ a - \frac{9a}{2} + 6a + d = 8 \Rightarrow \frac{5a}{2} + d = 8 \Rightarrow d = 8 - \frac{5a}{2}. \] and \[ 8a + 4\left(-\frac{9a}{2}\right) + 2(6a) + d = 4 \Rightarrow 2a + d = 4. \] Substitute \(d = 8 - \frac{5a}{2}\): \[ 2a + 8 - \frac{5a}{2} = 4 \Rightarrow -\frac{a}{2} = -4 \Rightarrow a = 8. \]
Now, \[ b = -\frac{9a}{2} = -36, \quad c = 48, \quad d = 8 - \frac{5(8)}{2} = -12. \]
Therefore, $$p(0) = d = -12.$$
Final Answer: $$\boxed{p(0) = -12}$$