Express the answer in simplest radical form. 2xsqrt [3](162x^10y^5) Answer Attemptiout of 2 square J.

To simplify the expression $2x\sqrt [3]{162x^{10}y^{5}}$, we need to simplify the cube root of $162x^{10}y^{5}$.<br /><br />First, let's simplify the cube root of $162$. We can rewrite $162$ as $2 \times 3^4$. Since $3^4$ is a perfect cube, we can take out $3^{4/3}$ (which is equal to $3^{1+1/3}$) from the cube root. So, $\sqrt[3]{162} = 2 \times 3^{4/3}$.<br /><br />Next, let's simplify the cube root of $x^{10}$. We can rewrite $x^{10}$ as $(x^3)^3 \times x$. Since $(x^3)^3$ is a perfect cube, we can take out $x^{9}$ from the cube root. So, $\sqrt[3]{x^{10}} = x^{9} \times \sqrt[3]{x}$.<br /><br />Finally, let's simplify the cube root of $y^{5}$. We can rewrite $y^{5}$ as $(y^2)^2 \times y$. Since $(y^2)^2$ is a perfect square, we can take out $y^2$ from the cube root. So, $\sqrt[3]{y^{5}} = y^2 \times \sqrt[3]{y}$.<br /><br />Putting it all together, we have:<br /><br />$2x\sqrt [3]{162x^{10}y^{5}} = 2x \times 2 \times 3^{4/3} \times x^{9} \times \sqrt[3]{x} \times y^2 \times \sqrt[3]{y}$<br /><br />Simplifying further, we get:<br /><br />$2x\sqrt [3]{162x^{10}y^{5}} = 4x \times 3^{4/3} \times x^{9} \times \sqrt[3]{x} \times y^2 \times \sqrt[3]{y}$<br /><br />Therefore, the simplified form of $2x\sqrt [3]{162x^{10}y^{5}}$ is $4x \times 3^{4/3} \times x^{9} \times \sqrt[3]{x} \times y^2 \times \sqrt[3]{y}$.