Question

use the product rule to simplify.
$7\sqrt3{81x^{4}y^{5}}$
$7\sqrt3{81x^{4}y^{5}} = \square$
(simplify your answer. type an exact answer, using radicals as needed.)

Explanation:

Step1: Factor the radicand

First, we factor the number inside the cube root and the variables. We know that \(81 = 27\times3\), \(x^{4}=x^{3}\times x\), and \(y^{5}=y^{3}\times y^{2}\). So we can rewrite the radicand as follows:
\(\sqrt[3]{81x^{4}y^{5}}=\sqrt[3]{27\times3\times x^{3}\times x\times y^{3}\times y^{2}}\)

Step2: Apply the product rule for radicals

The product rule for radicals states that \(\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}\) (where \(a\geq0\), \(b\geq0\) for real - valued roots). Using this rule, we can split the cube root:
\(\sqrt[3]{27\times3\times x^{3}\times x\times y^{3}\times y^{2}}=\sqrt[3]{27}\times\sqrt[3]{x^{3}}\times\sqrt[3]{y^{3}}\times\sqrt[3]{3xy^{2}}\)

Step3: Simplify the perfect cube roots

We know that \(\sqrt[3]{27} = 3\) (since \(3^{3}=27\)), \(\sqrt[3]{x^{3}}=x\) (by the property \(\sqrt[n]{a^{n}}=a\) for real \(a\) when \(n\) is odd), and \(\sqrt[3]{y^{3}} = y\). Substituting these values in, we get:
\(\sqrt[3]{27}\times\sqrt[3]{x^{3}}\times\sqrt[3]{y^{3}}\times\sqrt[3]{3xy^{2}}=3\times x\times y\times\sqrt[3]{3xy^{2}} = 3xy\sqrt[3]{3xy^{2}}\)

Step4: Multiply by the coefficient outside the radical

We have the original expression \(7\sqrt[3]{81x^{4}y^{5}}\). Substituting the simplified radical from Step 3, we get:
\(7\times3xy\sqrt[3]{3xy^{2}}=21xy\sqrt[3]{3xy^{2}}\)

Answer:

\(21xy\sqrt[3]{3xy^{2}}\)