what is the following product?\n$sqrt3{4}cdotsqrt{3}$\n$2(sqrt6{9})$\n$sqrt6{12}$\n$sqrt6{432}$\n$2(sqrt6{388…

Explanation:

Step1: Rewrite radicals with fractional exponents

The cube - root of 4 can be written as $4^{\frac{1}{3}}$, and the square - root of 3 can be written as $3^{\frac{1}{2}}$. So we have $4^{\frac{1}{3}}\times3^{\frac{1}{2}}$. Since $4 = 2^2$, then $4^{\frac{1}{3}}=(2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}}$ by the power - of - a - power rule $(a^m)^n=a^{mn}$. So the expression becomes $2^{\frac{2}{3}}\times3^{\frac{1}{2}}$.

Step2: Find a common denominator for the exponents

The common denominator of 3 and 2 is 6. Rewrite the exponents: $2^{\frac{2}{3}}=2^{\frac{4}{6}}$ and $3^{\frac{1}{2}}=3^{\frac{3}{6}}$. Then $2^{\frac{4}{6}}\times3^{\frac{3}{6}}=(2^4\times3^3)^{\frac{1}{6}}$ by the rule $a^m\times b^m=(a\times b)^m$.

Step3: Calculate the values inside the parentheses

Calculate $2^4 = 16$ and $3^3 = 27$. Then $2^4\times3^3=16\times27 = 432$. So $(2^4\times3^3)^{\frac{1}{6}}=\sqrt[6]{432}$.

Answer:

$\sqrt[6]{432}$ (corresponds to the third option)