Question

question 30 of 40
simplify the radical. assume that the variable represents a real number.
\\(\sqrt3{n^{16}}\\)
\\(\sqrt3{n^{16}} = \underline{\quad\quad}\\)

Explanation:

Step1: Recall the property of cube roots

For any real number \( a \) and integer \( m \), \( \sqrt[3]{a^m}=a^{\frac{m}{3}} \) when \( a \) is a real number. Also, we can express the exponent as a quotient and remainder. We need to divide the exponent of \( n \) (which is 16) by 3.

Step2: Divide 16 by 3

When we divide 16 by 3, we get \( 16 = 3\times5+1 \), so \( n^{16}=n^{3\times5 + 1}=(n^{3})^{5}\times n^{1} \) (using the property \( a^{m + n}=a^{m}\times a^{n} \) and \( a^{mn}=(a^{m})^{n} \)).

Step3: Simplify the cube root

\( \sqrt[3]{n^{16}}=\sqrt[3]{(n^{3})^{5}\times n}=\sqrt[3]{(n^{3})^{5}}\times\sqrt[3]{n} \) (using the property \( \sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b} \) for real numbers \( a,b \)). Since \( \sqrt[3]{(n^{3})^{5}} = n^{5} \) (because \( \sqrt[3]{a^{3}}=a \) and applying the power - of - a - power rule \( (a^{m})^{n}=a^{mn} \) and then the cube root property), we have \( \sqrt[3]{n^{16}}=n^{5}\sqrt[3]{n} \)

Answer:

\( n^{5}\sqrt[3]{n} \)