what is the simplest form of $sqrt4{324x^{6}y^{8}}$?\n$3xy^{2}sqrt4{2x^{2}}$\n$3xy^{2}sqrt4{4x^{2}}$\n$6xysqr…

Explanation:

Step1: Factorize the coefficient and variables

First, factorize 324 as (324 = 2\times162=2\times2\times81 = 2\times2\times3^4), (x^6=x^{4 + 2}=x^4\times x^2) and (y^8=(y^2)^4). So (\sqrt[4]{324x^{6}y^{8}}=\sqrt[4]{2\times2\times3^4\times x^4\times x^2\times(y^2)^4}).

Step2: Apply the fourth - root property (\sqrt[4]{ab}=\sqrt[4]{a}\cdot\sqrt[4]{b})

(\sqrt[4]{2\times2\times3^4\times x^4\times x^2\times(y^2)^4}=\sqrt[4]{3^4}\cdot\sqrt[4]{(y^2)^4}\cdot\sqrt[4]{x^4}\cdot\sqrt[4]{2\times2\times x^2}).

Step3: Simplify each fourth - root

We know that (\sqrt[4]{3^4}=3), (\sqrt[4]{(y^2)^4}=y^2), (\sqrt[4]{x^4}=x), and (\sqrt[4]{2\times2\times x^2}=\sqrt[4]{4x^2}). Then (\sqrt[4]{324x^{6}y^{8}} = 3xy^2\sqrt[4]{4x^2}).

Answer:

B. (3xy^{2}\sqrt[4]{4x^{2}})