1. **Simplify** $[(2a^3b^3 + 3a^2c)(4a^2b^3)]^2$. First, distribute inside the parentheses: $$ (2a^3b^3)(4a^2b^3) + (3a^2c)(4a^2b^3) = 8a^{3+2}b^{3+3} + 12a^{2+2}cb^3 = 8a^5b^6 + 12a^4cb^3 $$ Now square the entire expression: $$ (8a^5b^6 + 12a^4cb^3)^2 $$ Use the formula $(x + y)^2 = x^2 + 2xy + y^2$: $$ = (8a^5b^6)^2 + 2(8a^5b^6)(12a^4cb^3) + (12a^4cb^3)^2 $$ Calculate each term: $$ = 64a^{10}b^{12} + 192a^{9}b^{9}c + 144a^{8}b^{6}c^{2} $$ 2. **Simplify** $(2x^3 + 3)(3x^2 y)$. Distribute: $$ 2x^3 imes 3x^2 y + 3 imes 3x^2 y = 6x^{3+2}y + 9x^2 y = 6x^5 y + 9x^2 y $$ 3. **Simplify** $( oot{5}{x^2})(1 / x^{-2})$. Rewrite: $$ x^{\frac{2}{5}} \times x^{2} = x^{\frac{2}{5} + 2} = x^{\frac{2}{5} + \frac{10}{5}} = x^{\frac{12}{5}} $$ 4. **Simplify** $\sqrt[3]{x^4} \div \left(\frac{1}{x^4}\right)$. Rewrite division as multiplication: $$ x^{\frac{4}{3}} \times x^{4} = x^{\frac{4}{3} + 4} = x^{\frac{4}{3} + \frac{12}{3}} = x^{\frac{16}{3}} $$ **Final answers:** (a) $64a^{10}b^{12} + 192a^{9}b^{9}c + 144a^{8}b^{6}c^{2}$ (b) $6x^{5}y + 9x^{2}y$ (c) $x^{\frac{12}{5}}$ (d) $x^{\frac{16}{3}}$