Question

simplify \sqrt3{\frac{125x^{11}}{4}}. the simplified expression is \square.

Explanation:

Step1: Analyze the cube root of numerator and denominator

We know that for a cube root \(\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\) (where \(b
eq0\)). So we can split the given expression \(\sqrt[3]{\frac{125x^{11}}{4}}\) into \(\frac{\sqrt[3]{125x^{11}}}{\sqrt[3]{4}}\).

Step2: Simplify the cube root of the numerator

First, factorize \(125\) and \(x^{11}\). We know that \(125 = 5^3\) and \(x^{11}=x^{9 + 2}=x^{9}\cdot x^{2}=(x^{3})^{3}\cdot x^{2}\).
So \(\sqrt[3]{125x^{11}}=\sqrt[3]{5^{3}\cdot(x^{3})^{3}\cdot x^{2}}\).
Using the property \(\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}\) (for real numbers \(a,b\)), we get \(\sqrt[3]{5^{3}}\cdot\sqrt[3]{(x^{3})^{3}}\cdot\sqrt[3]{x^{2}} = 5x^{3}\sqrt[3]{x^{2}}\).

Step3: Rationalize the denominator

The denominator is \(\sqrt[3]{4}\), and we want to rationalize it. We know that \(4 = 2^{2}\), and to make the denominator a perfect cube, we multiply the numerator and denominator by \(\sqrt[3]{2}\) because \(\sqrt[3]{4}\cdot\sqrt[3]{2}=\sqrt[3]{4\times2}=\sqrt[3]{8} = 2\).
So \(\frac{5x^{3}\sqrt[3]{x^{2}}}{\sqrt[3]{4}}\times\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{5x^{3}\sqrt[3]{2x^{2}}}{\sqrt[3]{8}}=\frac{5x^{3}\sqrt[3]{2x^{2}}}{2}\).

Answer:

\(\dfrac{5x^{3}\sqrt[3]{2x^{2}}}{2}\)