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

Explanation:

Step1: Use 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 exponents of like - bases

$9x^{4}\cdot3x^{8}=(9\times3)x^{4 + 8}=27x^{12}$

Step3: Simplify the cube - root

$\sqrt[3]{27x^{12}}=\sqrt[3]{27}\cdot\sqrt[3]{x^{12}}$ Since $\sqrt[3]{27}=3$ and $\sqrt[3]{x^{12}}=x^{4}$ (because $12\div3 = 4$), we have $\sqrt[3]{27x^{12}}=3x^{4}$

Answer:

$3x^{4}$