1. **Problem Statement:** Find the intervals where the function $f(x) = (x+1)^3 (x-3)^3$ is strictly increasing or strictly decreasing. 2. **Formula and Rules:** To determine increasing or decreasing intervals, find the first derivative $f'(x)$ and analyze its sign. 3. **Find the derivative:** $$f(x) = (x+1)^3 (x-3)^3$$ Using the product rule: $$f'(x) = 3(x+1)^2 (x-3)^3 + (x+1)^3 3(x-3)^2$$ 4. **Factor the derivative:** $$f'(x) = 3(x+1)^2 (x-3)^2 \big[(x-3) + (x+1)\big] = 3(x+1)^2 (x-3)^2 (2x - 2)$$ 5. **Simplify:** $$f'(x) = 3(x+1)^2 (x-3)^2 2(x - 1) = 6(x+1)^2 (x-3)^2 (x - 1)$$ 6. **Critical points:** Set $f'(x) = 0$: $$(x+1)^2 = 0 \Rightarrow x = -1$$ $$(x-3)^2 = 0 \Rightarrow x = 3$$ $$(x-1) = 0 \Rightarrow x = 1$$ 7. **Sign analysis:** - $(x+1)^2$ and $(x-3)^2$ are always non-negative and zero only at $x=-1$ and $x=3$ respectively. - The sign of $f'(x)$ depends on $(x-1)$. 8. **Intervals:** - For $x < 1$, $(x-1) < 0$ so $f'(x) < 0$ (since other factors are non-negative). - For $x > 1$, $(x-1) > 0$ so $f'(x) > 0$. 9. **Behavior at critical points:** - At $x = -1$ and $x = 3$, derivative is zero but since these are squared factors, the sign of $f'(x)$ does not change. - At $x=1$, $f'(x)$ changes from negative to positive, indicating a local minimum. 10. **Conclusion:** - $f(x)$ is strictly decreasing on $(-\infty, 1)$. - $f(x)$ is strictly increasing on $(1, \infty)$. **Final answer:** $$\text{Increasing on } (1, \infty), \quad \text{Decreasing on } (-\infty, 1)$$