1. **State the problem:** Simplify the expression $$-2x^4y\sqrt{16y^{13}} - 2^{3}\sqrt{16x^{12}y^{16}} - x^{2}y^{3}\sqrt[3]{-54x^{6}y^{7}}$$ 2. **Recall the rules:** - For square roots, $$\sqrt{a^2} = a$$ if $$a \geq 0$$. - For cube roots, $$\sqrt[3]{a^3} = a$$ for any real $$a$$. - When simplifying radicals, factor inside the root to extract perfect powers. 3. **Simplify each radical:** - $$\sqrt{16y^{13}} = \sqrt{16} \cdot \sqrt{y^{12}} \cdot \sqrt{y} = 4y^{6}\sqrt{y}$$ - $$2^{3}\sqrt{16x^{12}y^{16}} = 8 \cdot \sqrt{16} \cdot \sqrt{x^{12}} \cdot \sqrt{y^{16}} = 8 \cdot 4 \cdot x^{6} \cdot y^{8} = 32x^{6}y^{8}$$ - $$\sqrt[3]{-54x^{6}y^{7}} = \sqrt[3]{-27 \cdot 2 \cdot x^{6} \cdot y^{6} \cdot y} = \sqrt[3]{-27} \cdot \sqrt[3]{2} \cdot \sqrt[3]{x^{6}} \cdot \sqrt[3]{y^{6}} \cdot \sqrt[3]{y} = -3 \cdot \sqrt[3]{2} \cdot x^{2} \cdot y^{2} \cdot \sqrt[3]{y} = -3x^{2}y^{2}\sqrt[3]{2y}$$ 4. **Substitute back:** $$-2x^{4}y \cdot 4y^{6}\sqrt{y} - 32x^{6}y^{8} - x^{2}y^{3} \cdot (-3x^{2}y^{2}\sqrt[3]{2y})$$ 5. **Multiply terms:** - $$-2x^{4}y \cdot 4y^{6}\sqrt{y} = -8x^{4}y^{7}\sqrt{y}$$ - $$- x^{2}y^{3} \cdot (-3x^{2}y^{2}\sqrt[3]{2y}) = +3x^{4}y^{5}\sqrt[3]{2y}$$ 6. **Write the full simplified expression:** $$-8x^{4}y^{7}\sqrt{y} - 32x^{6}y^{8} + 3x^{4}y^{5}\sqrt[3]{2y}$$ 7. **Final answer:** $$\boxed{-8x^{4}y^{7}\sqrt{y} - 32x^{6}y^{8} + 3x^{4}y^{5}\sqrt[3]{2y}}$$