Question

Help plsssss!!!!!! With this question!!!!

Answer 1

`\sqrt[3]{27}`

Explanation:

Let's split the cube root of 27 to make this easier:

`\sqrt[3]{27}=\sqrt[3]{3\cdot9}=\sqrt[3]{3\cdot3\cdot3}\\\text{Now, let's put }3\cdot3\cdot3 \text{ in exponential form. We would have:}\\\sqrt[3]{3^3}\\\text{Now, the 3 and 3 and the square root symbol would cancel out, and we }\\\text{ would be left with 3. And since three is an integer, we know that} } √(27) \text{ is perfect }`