QUESTION IMAGE
Question
rationalize the denominator of \\(\sqrt4{dfrac{256}{9x^3}}\\). assume that all variables represent positive real numbers.
Step1: Rewrite the radical as a fraction of radicals
We know that \(\sqrt[4]{\frac{a}{b}}=\frac{\sqrt[4]{a}}{\sqrt[4]{b}}\), so \(\sqrt[4]{\frac{256}{9x^{3}}}=\frac{\sqrt[4]{256}}{\sqrt[4]{9x^{3}}}\)
Since \(256 = 4^4\), then \(\sqrt[4]{256}=4\), so we have \(\frac{4}{\sqrt[4]{9x^{3}}}\)
Step2: Determine the missing factors to rationalize the denominator
For a fourth - root, we want the radicand in the denominator to be a perfect fourth - power. The radicand in the denominator is \(9x^{3}=3^{2}x^{3}\). We need to multiply the numerator and denominator by \(\sqrt[4]{3^{2}x}\) (because \((3^{2}x^{3})\times(3^{2}x)=3^{4}x^{4}\), which is a perfect fourth - power).
Step3: Multiply the numerator and denominator
Multiply the numerator and denominator by \(\sqrt[4]{9x}\) (note that \(3^{2}=9\)):
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Step4: Simplify the denominator
Since \(\sqrt[4]{81x^{4}}=\sqrt[4]{3^{4}x^{4}} = 3x\) (because \(x>0\))
Step5: Write the final expression
The expression becomes \(\frac{4\sqrt[4]{9x}}{3x}\)
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\(\frac{4\sqrt[4]{9x}}{3x}\)