Question

express in simplest radical form.
\\(x^2 y sqrt3{8 x^{11} y^4}\\)

Explanation:

Step1: Simplify the cube root term

First, rewrite the radicand by breaking down coefficients and variables into factors that are perfect cubes:
$$\sqrt[3]{8x^{11}y^4} = \sqrt[3]{8 \cdot x^9 \cdot x^2 \cdot y^3 \cdot y}$$
Since $8=2^3$, $x^9=(x^3)^3$, $y^3=(y)^3$, we can extract these perfect cubes:
$$\sqrt[3]{2^3 x^9 y^3 \cdot x^2 y} = 2x^3 y \sqrt[3]{x^2 y}$$

Step2: Multiply with the external terms

Multiply the simplified cube root with the $x^2 y$ term outside the radical, using exponent rules $x^a \cdot x^b = x^{a+b}$ and $y^a \cdot y^b = y^{a+b}$:
$$x^2 y \cdot 2x^3 y \sqrt[3]{x^2 y} = 2x^{2+3} y^{1+1} \sqrt[3]{x^2 y}$$
$$= 2x^5 y^2 \sqrt[3]{x^2 y}$$

Answer:

$2x^5 y^2 \sqrt[3]{x^2 y}$