Question

in order to pass the class, you must get at least 60%.
when you pass this exam, be sure to do the course exit ticket to get credit.
question 8
1 pts
simplify: $sqrt3{81x^{12}y^{15}}$
$9x^4y^3$
$3x^5y^5sqrt3{3}$
$3x^4y^5sqrt3{3}$
$9x^5y^3$

Explanation:

Step1: Factor the radicand

We can rewrite \(81\) as \(27\times3\), \(x^{12}\) as \((x^{4})^{3}\), and \(y^{15}\) as \((y^{5})^{3}\). So, \(\sqrt[3]{81x^{12}y^{15}}=\sqrt[3]{27\times3\times(x^{4})^{3}\times(y^{5})^{3}}\).

Step2: Use the property of cube roots

The property of cube roots states that \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) and \(\sqrt[3]{a^{3}} = a\). Applying these properties, we get:
\(\sqrt[3]{27\times3\times(x^{4})^{3}\times(y^{5})^{3}}=\sqrt[3]{27}\times\sqrt[3]{(x^{4})^{3}}\times\sqrt[3]{(y^{5})^{3}}\times\sqrt[3]{3}\)
Since \(\sqrt[3]{27} = 3\), \(\sqrt[3]{(x^{4})^{3}}=x^{4}\), and \(\sqrt[3]{(y^{5})^{3}} = y^{5}\), we substitute these values in:
\(3\times x^{4}\times y^{5}\times\sqrt[3]{3}=3x^{4}y^{5}\sqrt[3]{3}\)

Answer:

\(3x^{4}y^{5}\sqrt[3]{3}\) (corresponding to the option: \(3x^{4}y^{5}\sqrt[3]{3}\))