1. **State the problem:** Simplify the expression $$\sqrt[3]{40x^{4}y^{10}} + \left(125x^{15}y^{6}\right)^{\frac{1}{5}}$$ assuming all variables are positive. 2. **Recall the rules:** - The cube root of a product is the product of the cube roots: $$\sqrt[3]{a b} = \sqrt[3]{a} \cdot \sqrt[3]{b}$$. - For variables with exponents, $$\sqrt[n]{x^{m}} = x^{\frac{m}{n}}$$. - The fifth root of a power: $$\left(a^{m}\right)^{\frac{1}{5}} = a^{\frac{m}{5}}$$. - Since variables are positive, we do not need absolute values. 3. **Simplify the cube root term:** $$\sqrt[3]{40x^{4}y^{10}} = \sqrt[3]{40} \cdot \sqrt[3]{x^{4}} \cdot \sqrt[3]{y^{10}}$$ - Factor 40: $$40 = 8 \times 5$$, and $$\sqrt[3]{8} = 2$$. - So, $$\sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \cdot \sqrt[3]{5} = 2 \sqrt[3]{5}$$. - For $$x^{4}$$: $$\sqrt[3]{x^{4}} = x^{\frac{4}{3}}$$. - For $$y^{10}$$: $$\sqrt[3]{y^{10}} = y^{\frac{10}{3}}$$. Thus, $$\sqrt[3]{40x^{4}y^{10}} = 2 \sqrt[3]{5} \cdot x^{\frac{4}{3}} \cdot y^{\frac{10}{3}}$$. 4. **Simplify the fifth root term:** $$\left(125x^{15}y^{6}\right)^{\frac{1}{5}} = \left(125\right)^{\frac{1}{5}} \cdot \left(x^{15}\right)^{\frac{1}{5}} \cdot \left(y^{6}\right)^{\frac{1}{5}}$$ - Since $$125 = 5^{3}$$, $$\left(125\right)^{\frac{1}{5}} = 5^{\frac{3}{5}}$$. - For $$x^{15}$$: $$\left(x^{15}\right)^{\frac{1}{5}} = x^{3}$$. - For $$y^{6}$$: $$\left(y^{6}\right)^{\frac{1}{5}} = y^{\frac{6}{5}}$$. So, $$\left(125x^{15}y^{6}\right)^{\frac{1}{5}} = 5^{\frac{3}{5}} x^{3} y^{\frac{6}{5}}$$. 5. **Write the final simplified expression:** $$2 \sqrt[3]{5} x^{\frac{4}{3}} y^{\frac{10}{3}} + 5^{\frac{3}{5}} x^{3} y^{\frac{6}{5}}$$ This is the simplified form assuming all variables are positive.