simplify. (sqrt3{27x^{6}y^{3}}) assume all variables are nonnegative. enter your answer in the box.

Explanation:

Step1: Rewrite numbers as cubes

We know that $27 = 3^3$, so $\sqrt[3]{27x^{6}y^{3}}=\sqrt[3]{3^{3}x^{6}y^{3}}$.

Step2: Apply cube - root property

According to the property $\sqrt[3]{abc}=\sqrt[3]{a}\cdot\sqrt[3]{b}\cdot\sqrt[3]{c}$, we have $\sqrt[3]{3^{3}x^{6}y^{3}}=\sqrt[3]{3^{3}}\cdot\sqrt[3]{x^{6}}\cdot\sqrt[3]{y^{3}}$.

Step3: Simplify each cube - root

Since $\sqrt[3]{3^{3}} = 3$, $\sqrt[3]{x^{6}}=x^{2}$ (because $6\div3 = 2$) and $\sqrt[3]{y^{3}}=y$, then $\sqrt[3]{3^{3}}\cdot\sqrt[3]{x^{6}}\cdot\sqrt[3]{y^{3}}=3x^{2}y$.

Answer:

$3x^{2}y$