Solve the inequality algebraically. \[ 3 x^{3}>-24 x^{2} \] List the intervals and sign in each interval. Complete the following table. (Type your answers in interval notation. Use ascending order.) \begin{tabular}{|c|c|c|c|} \hline Interval & $(-8,0)$ & $U$ & $(0, \infty)$ \\ \hline Sign & Positive & Negative & Positive \\ \hline \end{tabular} What is the solution? (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression.)

Step 1 ：Set the inequality to 0: \(3x^3 + 24x^2 > 0\)

Step 2 ：Factor out a common factor of \(3x^2\): \(3x^2(x + 8) > 0\)

Step 3 ：Find the zeros of the inequality by setting each factor equal to 0: \(3x^2 = 0\) gives \(x = 0\) and \(x + 8 = 0\) gives \(x = -8\)

Step 4 ：Test a number from each interval in the inequality to determine the sign in that interval. For \((-\infty, -8)\), test \(x = -9\): \(3(-9)^2(-9 + 8) = -243\) which is less than 0. For \((-8, 0)\), test \(x = -1\): \(3(-1)^2(-1 + 8) = 21\) which is greater than 0. For \((0, \infty)\), test \(x = 1\): \(3(1)^2(1 + 8) = 27\) which is greater than 0.

Step 5 ：The solution to the inequality is where the expression is positive, which is in the intervals \((-8, 0)\) and \((0, \infty)\).

Step 6 ：\(\boxed{(-8, 0) \cup (0, \infty)}\) is the solution in interval notation.