For this case we must simplify the following expression:
[tex]\sqrt [3] {\frac {12x ^ 2} {16y}}[/tex]
We rewrite the expression as:
[tex]\sqrt[3]{\frac{4(3x^2)}{4(4y)}}=\\\sqrt[3]{\frac{4(3x^2)}{4(4y)}}=\\\frac{\sqrt[3]{3x^2}}{\sqrt[3]{4y}}=[/tex]
We multiply the numerator and denominator by:
[tex](\sqrt[3]{4y})^2:\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{\sqrt[3]{4y}*(\sqrt[3]{4y})^2}=[/tex]
We use the rule of power[tex]a ^ n * a ^ m = a ^ {n + m}[/tex] in the denominator:
[tex]\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{(\sqrt[3]{4y})^3}=\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{4y}=[/tex]
Move the exponent within the radical:
[tex]\frac{\sqrt[3]{3x^2}*(\sqrt[3]{16y^2}}{4y}=\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{2^3*(2y^2)}}{4y}=[/tex]
[tex]\frac{2\sqrt[3]{3x^2}*(\sqrt[3]{(2y^2)}}{4y}=\\\frac{2\sqrt[3]{6x^2*y^2}}{4y}=[/tex]
[tex]\frac{\sqrt[3]{6x^2*y^2}}{2y}[/tex]
Answer:
[tex]\frac{\sqrt[3]{6x^2*y^2}}{2y}[/tex]
