1. **State the problem:** Simplify the expression $$7x^3 + \frac{4}{x^5} - \frac{3}{\sqrt[3]{5x}} + \frac{7x^2}{3} - \sqrt[3]{5}.$$ 2. **Recall important rules:** - Negative exponents mean reciprocal: $$\frac{1}{x^n} = x^{-n}.$$ - Cube roots can be written as fractional exponents: $$\sqrt[3]{a} = a^{\frac{1}{3}}.$$ - Simplify terms by rewriting radicals and fractions as powers. 3. **Rewrite each term:** - $$7x^3$$ stays the same. - $$\frac{4}{x^5} = 4x^{-5}.$$ - $$\frac{3}{\sqrt[3]{5x}} = 3(5x)^{-\frac{1}{3}} = 3 \cdot 5^{-\frac{1}{3}} x^{-\frac{1}{3}}.$$ - $$\frac{7x^2}{3} = \frac{7}{3} x^2.$$ - $$\sqrt[3]{5} = 5^{\frac{1}{3}}.$$ 4. **Rewrite the expression:** $$7x^3 + 4x^{-5} - 3 \cdot 5^{-\frac{1}{3}} x^{-\frac{1}{3}} + \frac{7}{3} x^2 - 5^{\frac{1}{3}}.$$ 5. **Final simplified form:** $$7x^3 + 4x^{-5} - 3 \cdot 5^{-\frac{1}{3}} x^{-\frac{1}{3}} + \frac{7}{3} x^2 - 5^{\frac{1}{3}}.$$ This expression cannot be simplified further without specific values for $x$.