The problem asks to find the product of $3\sqrt{27}$ and $\sqrt{27}$ in simplest form and determine if the result is rational or irrational.

Step 1: Simplifying $\sqrt{27}$

$\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$

Step 2: Finding the product

We now need to compute the product $3\sqrt{27} \times \sqrt{27}$. Using the simplified form of $\sqrt{27}$, this becomes: $3\sqrt{27} \times \sqrt{27} = 3(3\sqrt{3}) \times 3\sqrt{3} = 9\sqrt{3} \times 3\sqrt{3}$ Next, we simplify the multiplication: $9\sqrt{3} \times 3\sqrt{3} = 9 \times 3 \times \sqrt{3} \times \sqrt{3} = 27 \times 3 = 81$

Step 3: Determining if the result is rational or irrational

Since the result is $81$, which is a whole number, it can be expressed as the ratio of two integers, $\frac{81}{1}$. Therefore, it is a rational number. Additionally, its decimal expansion terminates (it is just $81$).

Final Answer:

• Result: $81$
• The result is rational because it can be written as the ratio of two integers, and its decimal expansion terminates.

Let me know if you'd like any more details or clarifications.

Here are five related questions for you:

1. How do you simplify square roots of non-perfect squares?
2. What is the general rule for multiplying two square roots?
3. Can the square root of any positive integer be irrational?
4. What makes a number rational versus irrational?
5. How would you verify if a product of two expressions is rational or irrational?

Tip: Remember that the square of a square root eliminates the radical, turning it into the base number.