(i) Obtain the binomial expansion of `(2- sqrt(3) )^(6)` in the form `a+b sqrt(3)`, where `a and b` are integers. State the corresponding result for the expansion `( 2+ sqrt(3) )^(6)`

Answer

To find the binomial expansion of \( (2 - \sqrt{3})^6 \) in the form \( a + b\sqrt{3} \), we will use the Binomial Theorem, which states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In our case, we have \( a = 2 \), \( b = -\sqrt{3} \), and \( n = 6 \).

### Step 1: Write the Binomial Expansion

Using the Binomial Theorem, we can expand \( (2 - \sqrt{3})^6 \):

\[ (2 - \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-\sqrt{3})^k \]

### Step 2: Expand the Terms

Now we will calculate each term in the expansion:

1. For \( k = 0 \):
\[ \binom{6}{0} (2)^6 (-\sqrt{3})^0 = 1 \cdot 64 \cdot 1 = 64 \]

2. For \( k = 1 \):
\[ \binom{6}{1} (2)^5 (-\sqrt{3})^1 = 6 \cdot 32 \cdot (-\sqrt{3}) = -192\sqrt{3} \]

3. For \( k = 2 \):
\[ \binom{6}{2} (2)^4 (-\sqrt{3})^2 = 15 \cdot 16 \cdot 3 = 720 \]

4. For \( k = 3 \):
\[ \binom{6}{3} (2)^3 (-\sqrt{3})^3 = 20 \cdot 8 \cdot (-3\sqrt{3}) = -480\sqrt{3} \]

5. For \( k = 4 \):
\[ \binom{6}{4} (2)^2 (-\sqrt{3})^4 = 15 \cdot 4 \cdot 9 = 540 \]

6. For \( k = 5 \):
\[ \binom{6}{5} (2)^1 (-\sqrt{3})^5 = 6 \cdot 2 \cdot (-3\sqrt{3}) = -36\sqrt{3} \]

7. For \( k = 6 \):
\[ \binom{6}{6} (2)^0 (-\sqrt{3})^6 = 1 \cdot 1 \cdot 27 = 27 \]

### Step 3: Combine the Terms

Now, we will combine all the terms:

\[ (2 - \sqrt{3})^6 = 64 - 192\sqrt{3} + 720 - 480\sqrt{3} + 540 - 36\sqrt{3} + 27 \]

Combining the constant terms:

\[ 64 + 720 + 540 + 27 = 1351 \]

Combining the terms with \( \sqrt{3} \):

\[ -192\sqrt{3} - 480\sqrt{3} - 36\sqrt{3} = -708\sqrt{3} \]

### Final Result

Thus, the expansion of \( (2 - \sqrt{3})^6 \) is:

\[ (2 - \sqrt{3})^6 = 1351 - 708\sqrt{3} \]