Question

simplify:
\\(\frac{\sqrt3{5}}{\sqrt3{16x}}\\)
hint: be careful with parenthesis, be sure you use the preview button before hitting submit!!!)

Explanation:

Step1: Combine radicals into one

$\frac{\sqrt[3]{5}}{\sqrt[3]{16x}} = \sqrt[3]{\frac{5}{16x}}$

Step2: Simplify denominator inside radical

Factor $16 = 2^4 = 2^3 \times 2$, so $\sqrt[3]{\frac{5}{2^3 \times 2x}}$

Step3: Rationalize the radical

Multiply numerator and denominator inside by $2^2x^2$ to get a perfect cube in denominator:
$\sqrt[3]{\frac{5 \times 4x^2}{2^3 \times 2x \times 4x^2}} = \sqrt[3]{\frac{20x^2}{2^3 \times 8x^3}} = \sqrt[3]{\frac{20x^2}{(2 \times 2x)^3}} = \sqrt[3]{\frac{20x^2}{(4x)^3}}$

Step4: Split radical into numerator/denominator

$\frac{\sqrt[3]{20x^2}}{4x}$

Answer:

$\frac{\sqrt[3]{20x^2}}{4x}$