Question

rationalize the denominator of \\(\frac{6}{sqrt3{3}}\\). \\(\frac{6}{sqrt3{3}} = \square\\) (type an exact answer, using radicals as needed.)

Explanation:

Step1: Multiply by conjugate radical

Multiply numerator and denominator by $\sqrt[3]{3^2}$ to rationalize:
$$\frac{6}{\sqrt[3]{3}} \times \frac{\sqrt[3]{9}}{\sqrt[3]{9}} = \frac{6\sqrt[3]{9}}{\sqrt[3]{3 \times 9}}$$

Step2: Simplify the denominator

Simplify the cube root in the denominator:
$$\frac{6\sqrt[3]{9}}{\sqrt[3]{27}} = \frac{6\sqrt[3]{9}}{3}$$

Step3: Simplify the fraction

Divide the numerical coefficients:
$$\frac{6}{3}\sqrt[3]{9} = 2\sqrt[3]{9}$$

Answer:

$2\sqrt[3]{9}$