State, whether the following numbers are rational or not:
`(3- sqrt(3))^(2)`

Answer

To determine whether the number \((3 - \sqrt{3})^2\) is rational or irrational, we can follow these steps:

### Step 1: Expand the expression using the formula for the square of a binomial.
The formula for the square of a binomial \((a - b)^2\) is given by:
\[ (a - b)^2 = a^2 - 2ab + b^2 \]
In our case, let \(a = 3\) and \(b = \sqrt{3}\). Therefore, we can write:
\[ (3 - \sqrt{3})^2 = 3^2 - 2 \cdot 3 \cdot \sqrt{3} + (\sqrt{3})^2 \]

### Step 2: Calculate each term.
- \(3^2 = 9\)
- \((\sqrt{3})^2 = 3\)
- \(2 \cdot 3 \cdot \sqrt{3} = 6\sqrt{3}\)

### Step 3: Substitute the calculated values back into the expression.
Now substituting these values back, we have:
\[ (3 - \sqrt{3})^2 = 9 - 6\sqrt{3} + 3 \]

### Step 4: Combine like terms.
Combine the constant terms:
\[ 9 + 3 = 12 \]
Thus, we get:
\[ (3 - \sqrt{3})^2 = 12 - 6\sqrt{3} \]

### Step 5: Analyze the result.
Now we have \(12 - 6\sqrt{3}\). Here, \(12\) is a rational number, but \(6\sqrt{3}\) is an irrational number because \(\sqrt{3}\) is irrational.

### Step 6: Determine the nature of the entire expression.
The sum or difference of a rational number and an irrational number is always irrational. Therefore, since \(12\) is rational and \(6\sqrt{3}\) is irrational, the expression \(12 - 6\sqrt{3}\) is irrational.

### Conclusion:
Thus, \((3 - \sqrt{3})^2\) is an irrational number.

---