Question

simplify.
$sqrt5{160} - sqrt5{1215}$
$?sqrt5{}$

Explanation:

Step1: Simplify \( \sqrt[5]{160} \)

Factor 160 to find the 5th - power factor. We know that \( 160 = 32\times5 \), and \( 32 = 2^{5} \). So, by the property of radicals \( \sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b} \) (where \( n = 5 \), \( a = 2^{5} \), \( b = 5 \)), we have \( \sqrt[5]{160}=\sqrt[5]{2^{5}\times5}=2\sqrt[5]{5} \).

Step2: Simplify \( \sqrt[5]{1215} \)

Factor 1215. We find that \( 1215=243\times5 \), and \( 243 = 3^{5} \). Using the property of radicals \( \sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b} \) (where \( n = 5 \), \( a = 3^{5} \), \( b = 5 \)), we get \( \sqrt[5]{1215}=\sqrt[5]{3^{5}\times5}=3\sqrt[5]{5} \).

Step3: Subtract the two simplified radicals

Now, substitute the simplified forms into the original expression: \( \sqrt[5]{160}-\sqrt[5]{1215}=2\sqrt[5]{5}-3\sqrt[5]{5} \). Combining like terms (using the distributive property \( ac - bc=(a - b)c \), where \( a = 2 \), \( b = 3 \), \( c=\sqrt[5]{5} \)), we have \( (2 - 3)\sqrt[5]{5}=-1\sqrt[5]{5} \).

Answer:

\(-1\sqrt[5]{5}\) (or \(-\sqrt[5]{5}\))