If `a = 5+2 sqrt(6) and b = (1)/(a)`, then what will be the value of `a^(2) + b^(2) and a^(3) + b^(3)`?

Options

A. `98, 970`
B. `98, 1000`
C. `981, 985`
D. `970, 560`

Answer

To solve the problem, we need to find the values of \( a^2 + b^2 \) and \( a^3 + b^3 \) where \( a = 5 + 2\sqrt{6} \) and \( b = \frac{1}{a} \).

### Step 1: Calculate \( b \)

Given \( a = 5 + 2\sqrt{6} \), we can find \( b \) as follows:

\[ b = \frac{1}{a} = \frac{1}{5 + 2\sqrt{6}} \]

To simplify \( b \), we can multiply the numerator and denominator by the conjugate of the denominator:

\[ b = \frac{1 \cdot (5 - 2\sqrt{6})}{(5 + 2\sqrt{6})(5 - 2\sqrt{6})} \]

Calculating the denominator:

\[ (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1 \]

Thus, we have:

\[ b = 5 - 2\sqrt{6} \]

### Step 2: Calculate \( a^2 + b^2 \)

Now we can calculate \( a^2 \) and \( b^2 \):

\[ a^2 = (5 + 2\sqrt{6})^2 = 25 + 20\sqrt{6} + 24 = 49 + 20\sqrt{6} \]

\[ b^2 = (5 - 2\sqrt{6})^2 = 25 - 20\sqrt{6} + 24 = 49 - 20\sqrt{6} \]

Now, adding \( a^2 \) and \( b^2 \):

\[ a^2 + b^2 = (49 + 20\sqrt{6}) + (49 - 20\sqrt{6}) = 49 + 49 = 98 \]

### Step 3: Calculate \( a^3 + b^3 \)

We can use the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).

First, we need to find \( a + b \):

\[ a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10 \]

Next, we need to find \( ab \):

\[ ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 25 - 24 = 1 \]

Now we can find \( a^2 - ab + b^2 \):

\[ a^2 - ab + b^2 = (a^2 + b^2) - ab = 98 - 1 = 97 \]

Now substituting back into the identity:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 10 \times 97 = 970 \]

### Final Results

Thus, the values are:
- \( a^2 + b^2 = 98 \)
- \( a^3 + b^3 = 970 \)