Question

$sqrt3{9^2}$

Explanation:

Step1: Rewrite the radical expression

The expression is $\sqrt[3]{9^2}$. We can rewrite $9$ as $3^2$, so $9^2=(3^2)^2 = 3^{4}$. Wait, no, actually, $9 = 3^2$, so $9^2=(3^2)^2=3^{4}$? Wait, no, let's do it correctly. $9^2 = 81$, and we can also express $9$ as $3^2$, so $9^2=(3^2)^2 = 3^{4}$? Wait, no, $(a^m)^n=a^{mn}$, so $(3^2)^2 = 3^{2\times2}=3^4$. But we have a cube root. Alternatively, we can write $\sqrt[3]{9^2}$ as $9^{\frac{2}{3}}$, and since $9 = 3^2$, then $9^{\frac{2}{3}}=(3^2)^{\frac{2}{3}}=3^{\frac{4}{3}}=3^{1+\frac{1}{3}}=3\times3^{\frac{1}{3}} = 3\sqrt[3]{3}$. Wait, maybe another approach. Wait, the original expression is $\sqrt[3]{9^2}$, which is the cube root of $9$ squared. Let's compute $9^2 = 81$. Now, factor $81$: $81 = 27\times 3=3^3\times3$. So $\sqrt[3]{81}=\sqrt[3]{3^3\times3}=3\sqrt[3]{3}$.

Wait, let's check the steps again.

Step1: Express 9 as 3²

$9 = 3^2$, so $9^2=(3^2)^2$.

Step2: Simplify the exponent

Using the power of a power rule $(a^m)^n=a^{mn}$, we get $(3^2)^2 = 3^{2\times2}=3^4$. Wait, no, that's not helpful for the cube root. Alternatively, factor $9^2$ into prime factors. $9^2 = (3^2)^2 = 3^4$. Now, $3^4=3^{3 + 1}=3^3\times3^1$.

Step3: Apply the cube root

$\sqrt[3]{3^3\times3}=\sqrt[3]{3^3}\times\sqrt[3]{3}=3\times\sqrt[3]{3}=3\sqrt[3]{3}$.

Alternatively, we can write it as $9^{\frac{2}{3}}=(3^2)^{\frac{2}{3}}=3^{\frac{4}{3}}=3^{1+\frac{1}{3}}=3\times3^{\frac{1}{3}} = 3\sqrt[3]{3}$.

Answer:

$3\sqrt[3]{3}$ (or in exponent form $3^{\frac{4}{3}}$)