Question

simplify the radical expression.
\\(\sqrt5{x^{5}y^{5}}\\)
write your answer in the form \\(a, \sqrt5{b}\\), or \\(a\sqrt5{b}\\),
expressions in x and y. use at most one radical in your an
value symbol in your expression for a.

Explanation:

Step1: Recall the nth root property

For a non - negative real number \(a\) and positive integer \(n\), \(\sqrt[n]{a^{n}}=a\) when \(n\) is odd (and for all real \(a\) when \(n\) is odd, the \(n\)th root of \(a^{n}\) is \(a\)). Also, \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (product rule for radicals), where \(a\) and \(b\) are non - negative real numbers (when \(n\) is even) or any real numbers (when \(n\) is odd).

Given the radical expression \(\sqrt[5]{x^{5}y^{5}}\), we can use the product rule of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (here \(n = 5\), \(a=x^{5}\), \(b = y^{5}\)). So \(\sqrt[5]{x^{5}y^{5}}=\sqrt[5]{x^{5}}\cdot\sqrt[5]{y^{5}}\).

Step2: Apply the nth root property

Since \(n = 5\) (an odd number), by the property \(\sqrt[n]{a^{n}}=a\) (for any real number \(a\) when \(n\) is odd), we have \(\sqrt[5]{x^{5}}=x\) and \(\sqrt[5]{y^{5}}=y\).

Then \(\sqrt[5]{x^{5}}\cdot\sqrt[5]{y^{5}}=x\cdot y\).

Answer:

\(xy\)