1. **State the problem:** Find the product of the given expressions: (a) $\left(\sqrt{x^2 - y^2}\right)^7 \sqrt{(16x^2 - 16y^2)^2}$ (b) $(-2 - 5\sqrt{2})(-2 + 5\sqrt{2})$ (c) $\left(5 \sqrt[3]{m^2} - \sqrt[3]{4n}\right)\left(25m^3\sqrt[3]{m} + 5 \sqrt[3]{4m^2n} + \sqrt[3]{16n^2}\right)$ --- 2. **Recall formulas and rules:** - For radicals: $\sqrt{a^2} = |a|$ and $\sqrt[n]{a^m} = a^{m/n}$. - Difference of squares: $(a - b)(a + b) = a^2 - b^2$. - Cube roots and powers: $\sqrt[3]{a} = a^{1/3}$. - Distributive property for multiplication. --- 3. **Solve (a):** $$\left(\sqrt{x^2 - y^2}\right)^7 = (x^2 - y^2)^{7/2}$$ $$\sqrt{(16x^2 - 16y^2)^2} = |16x^2 - 16y^2| = 16|x^2 - y^2|$$ Assuming $x^2 - y^2 \geq 0$ for simplicity, then: $$16|x^2 - y^2| = 16(x^2 - y^2)$$ Multiply: $$(x^2 - y^2)^{7/2} \times 16(x^2 - y^2) = 16 (x^2 - y^2)^{7/2 + 1} = 16 (x^2 - y^2)^{9/2}$$ --- 4. **Solve (b):** Use difference of squares: $$(-2 - 5\sqrt{2})(-2 + 5\sqrt{2}) = (-2)^2 - (5\sqrt{2})^2 = 4 - 25 \times 2 = 4 - 50 = -46$$ --- 5. **Solve (c):** Let $a = 5 \sqrt[3]{m^2}$ and $b = \sqrt[3]{4n}$. Multiply: $$ (a - b)(25m^3\sqrt[3]{m} + 5 \sqrt[3]{4m^2n} + \sqrt[3]{16n^2}) $$ Rewrite terms using cube roots: - $25m^3\sqrt[3]{m} = 25 m^3 m^{1/3} = 25 m^{10/3}$ - $5 \sqrt[3]{4m^2n} = 5 (4 m^2 n)^{1/3} = 5 \sqrt[3]{4} m^{2/3} n^{1/3}$ - $\sqrt[3]{16n^2} = (16 n^2)^{1/3} = 16^{1/3} n^{2/3}$ Multiply out: $$a \times 25 m^{10/3} = 5 m^{2/3} \times 25 m^{10/3} = 125 m^{(2/3 + 10/3)} = 125 m^{12/3} = 125 m^4$$ $$a \times 5 \sqrt[3]{4} m^{2/3} n^{1/3} = 5 m^{2/3} \times 5 \sqrt[3]{4} m^{2/3} n^{1/3} = 25 \sqrt[3]{4} m^{4/3} n^{1/3}$$ $$a \times 16^{1/3} n^{2/3} = 5 m^{2/3} \times 16^{1/3} n^{2/3} = 5 \times 16^{1/3} m^{2/3} n^{2/3}$$ $$-b \times 25 m^{10/3} = - \sqrt[3]{4n} \times 25 m^{10/3} = -25 m^{10/3} \sqrt[3]{4n}$$ $$-b \times 5 \sqrt[3]{4} m^{2/3} n^{1/3} = -5 \sqrt[3]{4n} \times \sqrt[3]{4} m^{2/3} n^{1/3} = -5 m^{2/3} \sqrt[3]{4n \times 4 n} = -5 m^{2/3} \sqrt[3]{4^2 n^2} = -5 m^{2/3} \sqrt[3]{16 n^2}$$ $$-b \times 16^{1/3} n^{2/3} = - \sqrt[3]{4n} \times 16^{1/3} n^{2/3} = -16^{1/3} n^{2/3} \sqrt[3]{4n} = - \sqrt[3]{16 n^2} \sqrt[3]{4n} = - \sqrt[3]{64 n^3} = -4 n$$ Combine like terms: Notice $25 m^{10/3} \sqrt[3]{4n}$ and $-25 m^{10/3} \sqrt[3]{4n}$ cancel. Similarly, $25 \sqrt[3]{4} m^{4/3} n^{1/3}$ and $-5 m^{2/3} \sqrt[3]{16 n^2}$ do not cancel directly but are separate terms. Final simplified form: $$125 m^4 + 25 \sqrt[3]{4} m^{4/3} n^{1/3} + 5 \times 16^{1/3} m^{2/3} n^{2/3} - 5 m^{2/3} \sqrt[3]{16 n^2} - 4 n$$ --- **Final answers:** (a) $16 (x^2 - y^2)^{9/2}$ (b) $-46$ (c) $125 m^4 + 25 \sqrt[3]{4} m^{4/3} n^{1/3} + 5 \times 16^{1/3} m^{2/3} n^{2/3} - 5 m^{2/3} \sqrt[3]{16 n^2} - 4 n$