find the simplified product: $sqrt3{2x^{5}}cdotsqrt3{64x^{9}}$ \n$8x^{6}sqrt3{2x^{2}}$ \n$2x^{5}sqrt3{8x^{4}}…

Explanation:

Step1: Use the product - rule of radicals

$\sqrt[3]{2x^{5}}\cdot\sqrt[3]{64x^{9}}=\sqrt[3]{2x^{5}\cdot64x^{9}}$

Step2: Multiply the coefficients and add the exponents of $x$

$2\times64 = 128$ and $x^{5}\cdot x^{9}=x^{5 + 9}=x^{14}$, so we have $\sqrt[3]{128x^{14}}$

Step3: Rewrite the radicand

$128=64\times2$ and $x^{14}=x^{12}\cdot x^{2}$, so $\sqrt[3]{128x^{14}}=\sqrt[3]{64\times2\times x^{12}\times x^{2}}$

Step4: Use the product - rule of radicals again

$\sqrt[3]{64\times2\times x^{12}\times x^{2}}=\sqrt[3]{64}\cdot\sqrt[3]{x^{12}}\cdot\sqrt[3]{2x^{2}}$

Step5: Simplify each cube - root

$\sqrt[3]{64}=4$, $\sqrt[3]{x^{12}} = x^{4}$, so $\sqrt[3]{64}\cdot\sqrt[3]{x^{12}}\cdot\sqrt[3]{2x^{2}}=4x^{4}\sqrt[3]{2x^{2}}$

Answer:

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