1. **Stating the problem:** We want to verify or understand the integral equation: $$\int \frac{n^2}{\sqrt{27 - n^3}} \, du = -\frac{1}{2} t + c$$ 2. **Understanding the integral:** The integral is with respect to $u$, but the integrand is in terms of $n$. This suggests $n$ is a function of $u$, or there might be a substitution involved. 3. **Rewrite the integral:** Assuming $n$ is the variable of integration, the integral is: $$\int \frac{n^2}{\sqrt{27 - n^3}} \, dn$$ 4. **Use substitution:** Let: $$x = 27 - n^3$$ Then: $$\frac{dx}{dn} = -3n^2 \implies dx = -3n^2 dn$$ 5. **Rewrite $n^2 dn$ in terms of $dx$:** $$n^2 dn = -\frac{1}{3} dx$$ 6. **Substitute into the integral:** $$\int \frac{n^2}{\sqrt{27 - n^3}} dn = \int \frac{n^2}{\sqrt{x}} dn = \int \frac{1}{\sqrt{x}} \cdot n^2 dn = \int \frac{1}{\sqrt{x}} \cdot \left(-\frac{1}{3} dx\right) = -\frac{1}{3} \int x^{-1/2} dx$$ 7. **Integrate:** $$-\frac{1}{3} \int x^{-1/2} dx = -\frac{1}{3} \cdot 2 x^{1/2} + C = -\frac{2}{3} \sqrt{x} + C$$ 8. **Back-substitute $x$:** $$= -\frac{2}{3} \sqrt{27 - n^3} + C$$ 9. **Compare with the right side:** The given right side is: $$-\frac{1}{2} t + c$$ This suggests a relationship between $t$ and $n$ such that: $$-\frac{2}{3} \sqrt{27 - n^3} + C = -\frac{1}{2} t + c$$ or equivalently, $$t = \frac{4}{3} \sqrt{27 - n^3} + \text{constant}$$ **Final answer:** $$\int \frac{n^2}{\sqrt{27 - n^3}} \, dn = -\frac{2}{3} \sqrt{27 - n^3} + C$$