Question: 9^{3}\sqrt[3]{2w}(\sqrt[3]{4w} + 7^{3}\sqrt[3]{28}, w \geq 0

9^{3}\sqrt[3]{2w}(\sqrt[3]{4w} + 7^{3}\sqrt[3]{28}, w \geq 0

Solution

Let’s simplify the expression step by step. \[ 9\sqrt[3]{2w} \left( \sqrt[3]{4w} + 7\sqrt[3]{28} \right) \] First, simplify each cube root in the expression. \[ \sqrt[3]{4w} = \sqrt[3]{4} \cdot \sqrt[3]{w} = \sqrt[3]{2^2} \cdot \sqrt[3]{w} = 2^{2/3} w^{1/3} \] \[ \sqrt[3]{28} = \sqrt[3]{4 \cdot 7} = \sqrt[3]{4} \cdot \sqrt[3]{7} = 2^{2/3} 7^{1/3} \] Now, substitute these back into the original expression. \[ 9\sqrt[3]{2w} \left( 2^{2/3} w^{1/3} + 7 \cdot 2^{2/3} 7^{1/3} \right) \] Factor out the common term \(2^{2/3}\) from the terms inside the parentheses. \[ 9\sqrt[3]{2w} \cdot 2^{2/3} \left( w^{1/3} + 7 \cdot 7^{1/3} \right) \] Combine the constants outside the parentheses. \[ 9 \cdot 2^{2/3} \cdot \sqrt[3]{2w} \left( w^{1/3} + 7^{4/3} \right) \] Express \(\sqrt[3]{2w}\) as \(2^{1/3} w^{1/3}\). \[ 9 \cdot 2^{2/3} \cdot 2^{1/3} w^{1/3} \left( w^{1/3} + 7^{4/3} \right) \] Combine the powers of 2. \[ 9 \cdot 2^{(2/3 + 1/3)} w^{1/3} \left( w^{1/3} + 7^{4/3} \right) \] \[ 9 \cdot 2^{1} w^{1/3} \left( w^{1/3} + 7^{4/3} \right) \] \[ 18 w^{1/3} \left( w^{1/3} + 7^{4/3} \right) \] This is the simplified form of the original expression.