what is $\frac{sqrt3{3}}{sqrt3{2x}}$ in simplest form?\n$\frac{sqrt3{12x^{2}}}{2x}$\n$\frac{sqrt3{6x}}{2x}$\n…

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt[3]{2^{2}x^{2}}$. The original expression $\frac{\sqrt[3]{3}}{\sqrt[3]{2x}}$ becomes $\frac{\sqrt[3]{3}\times\sqrt[3]{2^{2}x^{2}}}{\sqrt[3]{2x}\times\sqrt[3]{2^{2}x^{2}}}$.

Step2: Use the property of cube - roots $\sqrt[3]{a}\times\sqrt[3]{b}=\sqrt[3]{ab}$

The numerator $\sqrt[3]{3}\times\sqrt[3]{2^{2}x^{2}}=\sqrt[3]{3\times4x^{2}}=\sqrt[3]{12x^{2}}$, and the denominator $\sqrt[3]{2x}\times\sqrt[3]{2^{2}x^{2}}=\sqrt[3]{2x\times4x^{2}}=\sqrt[3]{8x^{3}} = 2x$. So the expression is $\frac{\sqrt[3]{12x^{2}}}{2x}$.

Answer:

$\frac{\sqrt[3]{12x^{2}}}{2x}$