1. **Problem:** Evaluate $$\int \sqrt{x} (2\sqrt{x^3} - 1)^3 \, dx$$ by (a) expanding and (b) substitution. 2. **Step 1: Express the integrand clearly.** We have $$\sqrt{x} = x^{1/2}$$ and $$\sqrt{x^3} = (x^3)^{1/2} = x^{3/2}$$. 3. **Step 2: Expand $$(2\sqrt{x^3} - 1)^3$$ using binomial theorem:** $$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$ with $$a = 2x^{3/2}$$ and $$b = 1$$. 4. Calculate each term: $$a^3 = (2x^{3/2})^3 = 8x^{9/2}$$ $$3a^2b = 3(2x^{3/2})^2(1) = 3 \times 4x^3 = 12x^3$$ $$3ab^2 = 3(2x^{3/2})(1)^2 = 6x^{3/2}$$ $$b^3 = 1$$ 5. So, $$(2\sqrt{x^3} - 1)^3 = 8x^{9/2} - 12x^3 + 6x^{3/2} - 1$$ 6. **Step 3: Multiply by $$\sqrt{x} = x^{1/2}$$:** $$x^{1/2}(8x^{9/2} - 12x^3 + 6x^{3/2} - 1) = 8x^{5} - 12x^{7/2} + 6x^{2} - x^{1/2}$$ 7. **Step 4: Integrate term-by-term:** $$\int 8x^{5} dx = 8 \times \frac{x^{6}}{6} = \frac{4}{3}x^{6}$$ $$\int -12x^{7/2} dx = -12 \times \frac{x^{9/2}}{9/2} = -12 \times \frac{2}{9} x^{9/2} = -\frac{8}{3} x^{9/2}$$ $$\int 6x^{2} dx = 6 \times \frac{x^{3}}{3} = 2x^{3}$$ $$\int -x^{1/2} dx = - \frac{x^{3/2}}{3/2} = - \frac{2}{3} x^{3/2}$$ 8. **Step 5: Combine results:** $$\int \sqrt{x} (2\sqrt{x^3} - 1)^3 dx = \frac{4}{3}x^{6} - \frac{8}{3} x^{9/2} + 2x^{3} - \frac{2}{3} x^{3/2} + C$$ --- 9. **Step 6: Solve by substitution.** Let $$u = 2\sqrt{x^3} - 1 = 2x^{3/2} - 1$$. 10. Differentiate $$u$$: $$\frac{du}{dx} = 2 \times \frac{3}{2} x^{1/2} = 3x^{1/2}$$ 11. So, $$du = 3x^{1/2} dx \implies x^{1/2} dx = \frac{du}{3}$$ 12. Rewrite the integral: $$\int \sqrt{x} (2\sqrt{x^3} - 1)^3 dx = \int u^{3} \cdot x^{1/2} dx = \int u^{3} \frac{du}{3} = \frac{1}{3} \int u^{3} du$$ 13. Integrate: $$\frac{1}{3} \times \frac{u^{4}}{4} + C = \frac{u^{4}}{12} + C$$ 14. Substitute back: $$\frac{(2x^{3/2} - 1)^{4}}{12} + C$$ 15. **Final answers:** (a) $$\int \sqrt{x} (2\sqrt{x^3} - 1)^3 dx = \frac{4}{3}x^{6} - \frac{8}{3} x^{9/2} + 2x^{3} - \frac{2}{3} x^{3/2} + C$$ (b) $$\int \sqrt{x} (2\sqrt{x^3} - 1)^3 dx = \frac{(2x^{3/2} - 1)^{4}}{12} + C$$