find the simplified product: $sqrt3{9x^{4}}cdotsqrt3{3x^{8}}$ \n$sqrt3{12x^{12}}$ \n$sqrt3{27x^{12}}$…

Explanation:

Step1: Use the product - rule of radicals

$\sqrt[3]{9x^{4}}\cdot\sqrt[3]{3x^{8}}=\sqrt[3]{9x^{4}\cdot3x^{8}}$

Step2: Multiply the coefficients and add the exponents of like - bases

$9\times3 = 27$ and $x^{4}\cdot x^{8}=x^{4 + 8}=x^{12}$, so $\sqrt[3]{9x^{4}\cdot3x^{8}}=\sqrt[3]{27x^{12}}$

Step3: Simplify the cube - root

$\sqrt[3]{27x^{12}}=\sqrt[3]{27}\cdot\sqrt[3]{x^{12}}$. Since $\sqrt[3]{27}=3$ and $\sqrt[3]{x^{12}}=x^{4}$, then $\sqrt[3]{27x^{12}} = 3x^{4}$

Answer:

$3x^{4}$