1. **State the problem:** Simplify the expression $$2x \sqrt[3]{16x^{4}} + \sqrt[3]{375x^{8}} - 2x^{3} \sqrt[3]{54x}$$ and express it in the form $$Ax^{2} \sqrt[3]{Bx^{2}} - Cx^{n} \sqrt[3]{2x}$$. 2. **Recall the cube root and exponent rules:** - $$\sqrt[3]{a^{m}} = a^{\frac{m}{3}}$$ - When multiplying powers with the same base, add exponents: $$a^{m} \cdot a^{n} = a^{m+n}$$ - When factoring inside cube roots, separate perfect cubes to simplify. 3. **Simplify each term:** - First term: $$2x \sqrt[3]{16x^{4}} = 2x \sqrt[3]{16} \sqrt[3]{x^{4}}$$ Since $$16 = 2^{4}$$, and $$\sqrt[3]{2^{4}} = 2^{\frac{4}{3}}$$, and $$\sqrt[3]{x^{4}} = x^{\frac{4}{3}}$$, so first term becomes: $$2x \cdot 2^{\frac{4}{3}} \cdot x^{\frac{4}{3}} = 2 \cdot 2^{\frac{4}{3}} \cdot x^{1 + \frac{4}{3}} = 2^{1 + \frac{4}{3}} x^{\frac{7}{3}} = 2^{\frac{7}{3}} x^{\frac{7}{3}}$$ - Second term: $$\sqrt[3]{375x^{8}} = \sqrt[3]{375} \cdot \sqrt[3]{x^{8}}$$ Factor 375: $$375 = 125 \times 3 = 5^{3} \times 3$$, so $$\sqrt[3]{375} = \sqrt[3]{5^{3} \times 3} = 5 \sqrt[3]{3}$$ and $$\sqrt[3]{x^{8}} = x^{\frac{8}{3}}$$ So second term is: $$5 \sqrt[3]{3} x^{\frac{8}{3}}$$ - Third term: $$-2x^{3} \sqrt[3]{54x} = -2x^{3} \sqrt[3]{54} \sqrt[3]{x}$$ Factor 54: $$54 = 27 \times 2 = 3^{3} \times 2$$, so $$\sqrt[3]{54} = \sqrt[3]{3^{3} \times 2} = 3 \sqrt[3]{2}$$ and $$\sqrt[3]{x} = x^{\frac{1}{3}}$$ So third term is: $$-2x^{3} \cdot 3 \sqrt[3]{2} \cdot x^{\frac{1}{3}} = -6 x^{3 + \frac{1}{3}} \sqrt[3]{2} = -6 x^{\frac{10}{3}} \sqrt[3]{2}$$ 4. **Rewrite all terms with fractional exponents:** - First term: $$2^{\frac{7}{3}} x^{\frac{7}{3}}$$ - Second term: $$5 x^{\frac{8}{3}} \sqrt[3]{3}$$ - Third term: $$-6 x^{\frac{10}{3}} \sqrt[3]{2}$$ 5. **Express terms to match the form $$Ax^{2} \sqrt[3]{Bx^{2}} - Cx^{n} \sqrt[3]{2x}$$:** - For the first term, write $$x^{\frac{7}{3}} = x^{2 + \frac{1}{3}} = x^{2} x^{\frac{1}{3}}$$ and $$2^{\frac{7}{3}} = 2^{2 + \frac{1}{3}} = 4 \cdot 2^{\frac{1}{3}}$$ So first term: $$2^{\frac{7}{3}} x^{\frac{7}{3}} = 4 \cdot 2^{\frac{1}{3}} x^{2} x^{\frac{1}{3}} = 4 x^{2} \sqrt[3]{2x}$$ - Second term: $$5 x^{\frac{8}{3}} \sqrt[3]{3} = 5 x^{2 + \frac{2}{3}} \sqrt[3]{3} = 5 x^{2} x^{\frac{2}{3}} \sqrt[3]{3} = 5 x^{2} \sqrt[3]{3 x^{2}}$$ - Third term is already in the form $$-6 x^{\frac{10}{3}} \sqrt[3]{2} = -6 x^{3 + \frac{1}{3}} \sqrt[3]{2} = -6 x^{n} \sqrt[3]{2x}$$ with $$n = 3 + \frac{1}{3} = \frac{10}{3}$$ 6. **Combine the expression:** $$4 x^{2} \sqrt[3]{2x} + 5 x^{2} \sqrt[3]{3 x^{2}} - 6 x^{\frac{10}{3}} \sqrt[3]{2x}$$ 7. **Identify constants:** $$A = 5, B = 3, C = 6, n = \frac{10}{3}$$ Note the first term can be rewritten as $$4 x^{2} \sqrt[3]{2x}$$ which matches the $$- C x^{n} \sqrt[3]{2x}$$ form but with positive sign, so the expression is: $$5 x^{2} \sqrt[3]{3 x^{2}} + 4 x^{2} \sqrt[3]{2x} - 6 x^{\frac{10}{3}} \sqrt[3]{2x}$$ or grouping terms with $$\sqrt[3]{2x}$$: $$5 x^{2} \sqrt[3]{3 x^{2}} + (4 x^{2} - 6 x^{\frac{10}{3}}) \sqrt[3]{2x}$$ **Final answer:** $$2x \sqrt[3]{16x^{4}} + \sqrt[3]{375x^{8}} - 2x^{3} \sqrt[3]{54x} = 5 x^{2} \sqrt[3]{3 x^{2}} + 4 x^{2} \sqrt[3]{2x} - 6 x^{\frac{10}{3}} \sqrt[3]{2x}$$