Question

$\sqrt4{324x^{5}}$

Explanation:

Step1: Factor the radicand

First, factor \(324\) and \(x^5\) to find perfect fourth - powers. We know that \(324 = 81\times4=3^4\times4\) and \(x^5=x^{4 + 1}=x^4\times x\). So, \(\sqrt[4]{324x^{5}}=\sqrt[4]{3^4\times4\times x^4\times x}\)

Step2: Use the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\))

According to the property \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\), we can split the radical as follows:
\(\sqrt[4]{3^4\times4\times x^4\times x}=\sqrt[4]{3^4}\cdot\sqrt[4]{x^4}\cdot\sqrt[4]{4x}\)

Step3: Simplify the perfect fourth - powers

Since \(\sqrt[4]{a^4}=a\) for \(a\geq0\), we have \(\sqrt[4]{3^4} = 3\) and \(\sqrt[4]{x^4}=x\) (assuming \(x\geq0\)). Then the expression becomes \(3x\sqrt[4]{4x}\)

Answer:

\(3x\sqrt[4]{4x}\)