Question

simplify the expression, using absolute value signs as necessary.
(sqrt5{1215a^{17}m^{13}})
show your work here
hint to add the nth root symbol ((sqrtn{square})), type
oot\

Explanation:

Step1: Factor the radicand

First, factor each part of the radicand \(1215a^{17}m^{13}\) into perfect fifth - powers and remaining factors.

• For the coefficient: We know that \(1215 = 5\times243=5\times3^5\) (since \(3^5=243\)).
• For the variable \(a\): Using the property of exponents \(a^{mn}=(a^m)^n\), we can write \(a^{17}=a^{15 + 2}=a^{5\times3+2}=(a^3)^5\times a^2\).
• For the variable \(m\): \(m^{13}=m^{10 + 3}=m^{5\times2+3}=(m^2)^5\times m^3\).

So, \(\sqrt[5]{1215a^{17}m^{13}}=\sqrt[5]{3^5\times5\times(a^3)^5\times a^2\times(m^2)^5\times m^3}\).

Step2: Use the property of n - th roots \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) and \(\sqrt[n]{a^n}=a\) (for real numbers and \(n\) odd)

Using the property \(\sqrt[n]{abc\cdots}=\sqrt[n]{a}\cdot\sqrt[n]{b}\cdot\sqrt[n]{c}\cdots\), we can split the radical:
\(\sqrt[5]{3^5\times5\times(a^3)^5\times a^2\times(m^2)^5\times m^3}=\sqrt[5]{3^5}\cdot\sqrt[5]{(a^3)^5}\cdot\sqrt[5]{(m^2)^5}\cdot\sqrt[5]{5a^2m^3}\)

Since \(\sqrt[5]{x^5}=x\) for any real number \(x\) (because the fifth - root is an odd root), we have \(\sqrt[5]{3^5} = 3\), \(\sqrt[5]{(a^3)^5}=a^3\), \(\sqrt[5]{(m^2)^5}=m^2\).

So, the expression becomes \(3a^3m^2\sqrt[5]{5a^2m^3}\)

Answer:

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