what is the simplest form of $sqrt4{81x^{8}y^{5}}$?\n$3x^{2}(sqrt4{y^{5}})$\n$3x^{2}y(sqrt4{y})$\n$9x^{2}y(sq…

what is the simplest form of $sqrt4{81x^{8}y^{5}}$?\n$3x^{2}(sqrt4{y^{5}})$\n$3x^{2}y(sqrt4{y})$\n$9x^{2}y(sqrt4{y})$\n$9x^{4}y^{2}(sqrt4{y})$

what is the simplest form of $sqrt4{81x^{8}y^{5}}$?\n$3x^{2}(sqrt4{y^{5}})$\n$3x^{2}y(sqrt4{y})$\n$9x^{2}y(sqrt4{y})$\n$9x^{4}y^{2}(sqrt4{y})$

Answer

Explanation:

Step1: Simplify the coefficient

We know that $\sqrt[4]{81}=3$ since $3^4 = 81$.

Step2: Simplify the variable part with $x$

For $x^8$ in the fourth - root, using the rule $\sqrt[n]{a^m}=a^{\frac{m}{n}}$, we have $\sqrt[4]{x^8}=x^{\frac{8}{4}}=x^2$.

Step3: Simplify the variable part with $y$

For $y^5$ in the fourth - root, $\sqrt[4]{y^5}=y^{\frac{5}{4}}=y\cdot y^{\frac{1}{4}}=y\sqrt[4]{y}$.

Step4: Combine the results

$\sqrt[4]{81x^8y^5}=\sqrt[4]{81}\cdot\sqrt[4]{x^8}\cdot\sqrt[4]{y^5}=3x^2y\sqrt[4]{y}$.

Answer:

$3x^2y(\sqrt[4]{y})$