Solve the inequality algebraically. \[ x^{3}-2 x^{2}-63 x<0 \] The solution is (Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)

To solve the inequality \( x^{3}-2 x^{2}-63 x<0 \), we first factor the expression: 1. Factor out \( x \): \[ x(x^2 - 2x - 63) < 0 \] 2. Next, we need to factor the quadratic \( x^2 - 2x - 63 \). We look for two numbers that multiply to \(-63\) and add to \(-2\). The numbers \(-9\) and \(7\) work: \[ x^2 - 2x - 63 = (x - 9)(x + 7) \] 3. So the inequality becomes: \[ x(x - 9)(x + 7) < 0 \] 4. Now we find the critical points by setting the expression equal to zero: \[ x = 0, \quad x - 9 = 0 \Rightarrow x = 9, \quad x + 7 = 0 \Rightarrow x = -7 \] 5. The critical points are \( x = -7, 0, 9 \). We will test intervals defined by these points: \( (-\infty, -7) \), \( (-7, 0) \), \( (0, 9) \), and \( (9, \infty) \). 6. Choose a test point from each interval: - For \( x < -7 \) (e.g., \( x = -8 \)): \[ -8(-8 - 9)(-8 + 7) = -8(-17)(-1) > 0 \] - For \( -7 < x < 0 \) (e.g., \( x = -1 \)): \[ -1(-1 - 9)(-1 + 7) = -1(-10)(6) < 0 \] - For \( 0 < x < 9 \) (e.g., \( x = 1 \)): \[ 1(1 - 9)(1 + 7) = 1(-8)(8) < 0 \] - For \( x > 9 \) (e.g., \( x = 10 \)): \[ 10(10 - 9)(10 + 7) = 10(1)(17) > 0 \] 7. From the tests, the intervals where the inequality \( x(x - 9)(x + 7) < 0 \) holds are \( (-7, 0) \) and \( (0, 9) \). Thus, the solution in interval notation is: \[ \boxed{(-7, 0) \cup (0, 9)} \]