1. **State the problem:** Evaluate the integral $$\int 6 \sqrt[3]{x} \left( \frac{x^3}{3} + \frac{7}{2} x^{-\frac{1}{3}} \right) dx.$$\n\n2. **Rewrite the integral:** Recall that $$\sqrt[3]{x} = x^{\frac{1}{3}}.$$ So the integral becomes\n$$\int 6 x^{\frac{1}{3}} \left( \frac{x^3}{3} + \frac{7}{2} x^{-\frac{1}{3}} \right) dx.$$\n\n3. **Distribute inside the integral:**\n$$\int 6 \left( \frac{x^{\frac{1}{3} + 3}}{3} + \frac{7}{2} x^{\frac{1}{3} - \frac{1}{3}} \right) dx = \int 6 \left( \frac{x^{\frac{10}{3}}}{3} + \frac{7}{2} x^0 \right) dx.$$\n\n4. **Simplify constants:**\n$$\int \left( 2 x^{\frac{10}{3}} + 21 \right) dx.$$\n\n5. **Integrate term-by-term:**\n- For $$2 x^{\frac{10}{3}}$$, use the power rule: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C.$$\n- For $$21$$, integral is $$21x.$$\n\nSo,\n$$\int 2 x^{\frac{10}{3}} dx = 2 \cdot \frac{x^{\frac{10}{3} + 1}}{\frac{10}{3} + 1} = 2 \cdot \frac{x^{\frac{13}{3}}}{\frac{13}{3}} = 2 \cdot \frac{3}{13} x^{\frac{13}{3}} = \frac{6}{13} x^{\frac{13}{3}}.$$\n\n6. **Write the full integral result:**\n$$\frac{6}{13} x^{\frac{13}{3}} + 21x + C,$$ where $$C$$ is the constant of integration.\n\n**Final answer:** $$\int 6 \sqrt[3]{x} \left( \frac{x^3}{3} + \frac{7}{2} x^{-\frac{1}{3}} \right) dx = \frac{6}{13} x^{\frac{13}{3}} + 21x + C.$$