If m `= (1)/(3-2sqrt(2))` and `n = (1)/(3+2sqrt(2))` find :
(i) `m^(2)` (ii) `n^(2)` (iii) mn

Answer

To solve the problem, we need to find \( m^2 \), \( n^2 \), and \( mn \) given:

\[ m = \frac{1}{3 - 2\sqrt{2}} \quad \text{and} \quad n = \frac{1}{3 + 2\sqrt{2}} \]

### Step 1: Rationalizing \( m \)

To rationalize \( m \), we multiply the numerator and denominator by the conjugate of the denominator, which is \( 3 + 2\sqrt{2} \):

\[ m = \frac{1}{3 - 2\sqrt{2}} \cdot \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} \]

Calculating the denominator using the difference of squares:

\[ (3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]

Thus, we have:

\[ m = 3 + 2\sqrt{2} \]

### Step 2: Rationalizing \( n \)

Next, we rationalize \( n \) in a similar manner by multiplying by the conjugate \( 3 - 2\sqrt{2} \):

\[ n = \frac{1}{3 + 2\sqrt{2}} \cdot \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} \]

Calculating the denominator:

\[ (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]

Thus, we have:

\[ n = 3 - 2\sqrt{2} \]

### Step 3: Finding \( m^2 \)

Now, we calculate \( m^2 \):

\[ m^2 = (3 + 2\sqrt{2})^2 \]

Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):

\[ = 3^2 + 2 \cdot 3 \cdot 2\sqrt{2} + (2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \]

### Step 4: Finding \( n^2 \)

Now, we calculate \( n^2 \):

\[ n^2 = (3 - 2\sqrt{2})^2 \]

Using the same formula:

\[ = 3^2 - 2 \cdot 3 \cdot 2\sqrt{2} + (2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \]

### Step 5: Finding \( mn \)

Now we calculate \( mn \):

\[ mn = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \]

Using the difference of squares:

\[ = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]

### Final Results

Thus, the final results are:

(i) \( m^2 = 17 + 12\sqrt{2} \)

(ii) \( n^2 = 17 - 12\sqrt{2} \)

(iii) \( mn = 1 \)