If ` a = 7 + 4 sqrt( 3) `
find the value of `( 3 a ^(6) + 2a ^(4) + 4a^(3) + 2a^(2) + 3)/( a^(4) + a^(3) + a^(2))`

Options

A. `(8138)/(17)`
B. `(8138)/(15)`
C. `(8238)/(15)`
D. `(8338)/(15)`

Answer

To solve the expression \(\frac{3a^6 + 2a^4 + 4a^3 + 2a^2 + 3}{a^4 + a^3 + a^2}\) where \(a = 7 + 4\sqrt{3}\), we will follow these steps:

### Step 1: Calculate \( \frac{1}{a} \)

Given \( a = 7 + 4\sqrt{3} \), we can find \( \frac{1}{a} \) by rationalizing the denominator:

\[ \frac{1}{a} = \frac{1}{7 + 4\sqrt{3}} \cdot \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{7 - 4\sqrt{3}}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} \]

Calculating the denominator:

\[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \]

Thus,

\[ \frac{1}{a} = 7 - 4\sqrt{3} \]

### Step 2: Calculate \( a + \frac{1}{a} \)

Now we can find \( a + \frac{1}{a} \):

\[ a + \frac{1}{a} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]

### Step 3: Calculate \( a^3 + \frac{1}{a^3} \)

Using the identity:

\[ a^3 + \frac{1}{a^3} = (a + \frac{1}{a})^3 - 3(a + \frac{1}{a}) \]

Substituting \( a + \frac{1}{a} = 14 \):

\[ a^3 + \frac{1}{a^3} = 14^3 - 3 \cdot 14 \]

Calculating \( 14^3 \):

\[ 14^3 = 2744 \]

Now substituting:

\[ a^3 + \frac{1}{a^3} = 2744 - 42 = 2702 \]

### Step 4: Substitute into the original expression

Now we can substitute \( a^3 + \frac{1}{a^3} \) and \( a + \frac{1}{a} \) into the original expression:

\[ \frac{3a^6 + 2a^4 + 4a^3 + 2a^2 + 3}{a^4 + a^3 + a^2} \]

We can factor out \( a^3 \) from the numerator and denominator:

Numerator:

\[ 3a^6 + 2a^4 + 4a^3 + 2a^2 + 3 = a^3(3a^3 + 2a + 4) + 2a^2 + 3 \]

Denominator:

\[ a^4 + a^3 + a^2 = a^2(a^2 + a + 1) \]

### Step 5: Simplify the expression

Now we can simplify the expression:

\[ \frac{3a^3 + 2a + 4 + \frac{2}{a^3} + \frac{3}{a^3}}{a^2 + a + 1} \]

Substituting \( a + \frac{1}{a} = 14 \) and \( a^3 + \frac{1}{a^3} = 2702 \):

\[ = \frac{3(2702) + 2(14) + 4}{14 + 1} \]

Calculating:

\[ = \frac{8106 + 28 + 4}{15} = \frac{8138}{15} \]

### Final Answer

Thus, the final answer is:

\[ \frac{8138}{15} \]