what is the following product?\n$sqrt3{16x^{7}}cdotsqrt3{12x^{9}}$\n$x^{2}(sqrt3{28x^{2}})$\n$x^{5}(sqrt3{28x…

Explanation:

Step1: Use the product - rule of radicals

$\sqrt[3]{16x^{7}}\cdot\sqrt[3]{12x^{9}}=\sqrt[3]{16x^{7}\cdot12x^{9}}$

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

$16\times12 = 192$, and $x^{7}\cdot x^{9}=x^{7 + 9}=x^{16}$. So we have $\sqrt[3]{192x^{16}}$.

Step3: Prime - factorize 192 and rewrite the exponent of $x$

$192=2^{6}\times3$, and $x^{16}=x^{15}\cdot x$. Then $\sqrt[3]{192x^{16}}=\sqrt[3]{2^{6}\times3\times x^{15}\times x}$.

Step4: Use the property $\sqrt[3]{a\cdot b}=\sqrt[3]{a}\cdot\sqrt[3]{b}$

$\sqrt[3]{2^{6}\times3\times x^{15}\times x}=\sqrt[3]{2^{6}}\cdot\sqrt[3]{3}\cdot\sqrt[3]{x^{15}}\cdot\sqrt[3]{x}$.

Step5: Simplify each cube - root

$\sqrt[3]{2^{6}} = 2^{2}=4$, $\sqrt[3]{x^{15}}=x^{5}$. So $\sqrt[3]{2^{6}}\cdot\sqrt[3]{3}\cdot\sqrt[3]{x^{15}}\cdot\sqrt[3]{x}=4x^{5}\sqrt[3]{3x}$.

Answer:

$4x^{5}\left(\sqrt[3]{3x}\right)$