Question

write the following in simplified radical form.
\\(\sqrt5{160w^8}\\)
assume that the variable represents a positive real number.

Explanation:

Step1: Factor the radicand

First, factor \(160\) and \(w^8\) into parts that are perfect fifth - powers.
We know that \(160 = 32\times5\), and \(32 = 2^5\). Also, for the variable part, \(w^8=w^{5 + 3}=w^5\times w^3\).
So, \(\sqrt[5]{160w^8}=\sqrt[5]{2^5\times5\times w^5\times w^3}\)

Step2: Use the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}\) (\(a\geq0,b\geq0\) for even \(n\); \(a,b\in\mathbb{R}\) for odd \(n\), here \(n = 5\) which is odd)

\(\sqrt[5]{2^5\times5\times w^5\times w^3}=\sqrt[5]{2^5}\times\sqrt[5]{w^5}\times\sqrt[5]{5w^3}\)

Step3: Simplify the perfect fifth - power radicals

Since \(\sqrt[5]{2^5}=2\) and \(\sqrt[5]{w^5}=w\) (because \(n = 5\) and for any real number \(x\), \(\sqrt[n]{x^n}=x\) when \(n\) is odd)
So, \(\sqrt[5]{2^5}\times\sqrt[5]{w^5}\times\sqrt[5]{5w^3}=2\times w\times\sqrt[5]{5w^3}=2w\sqrt[5]{5w^3}\)

Answer:

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