The smallest integer larger than `(sqrt(3) + sqrt(2))^(6)` is

Options

A. `969`
B. `970`
C. `971`
D. `972`

Explanation

To find the smallest integer larger than \((\sqrt{3} + \sqrt{2})^6\), we can use the binomial theorem and some properties of binomial expansions. Here’s the step-by-step solution:

### Step 1: Use the Binomial Theorem
The binomial theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
We can apply this to our expression \((\sqrt{3} + \sqrt{2})^6\).

### Step 2: Set Up the Expression
Let \(a = \sqrt{3}\) and \(b = \sqrt{2}\), and \(n = 6\). Therefore, we have:
\[ (\sqrt{3} + \sqrt{2})^6 + (\sqrt{3} - \sqrt{2})^6 = 2 \left( \binom{6}{0} (\sqrt{3})^6 + \binom{6}{2} (\sqrt{3})^4 (\sqrt{2})^2 + \binom{6}{4} (\sqrt{3})^2 (\sqrt{2})^4 + \binom{6}{6} (\sqrt{2})^6 \right) \]

### Step 3: Calculate Each Term
Now we calculate each term:
- \(\binom{6}{0} (\sqrt{3})^6 = 1 \cdot 27 = 27\)
- \(\binom{6}{2} (\sqrt{3})^4 (\sqrt{2})^2 = 15 \cdot 9 \cdot 2 = 270\)
- \(\binom{6}{4} (\sqrt{3})^2 (\sqrt{2})^4 = 15 \cdot 3 \cdot 4 = 180\)
- \(\binom{6}{6} (\sqrt{2})^6 = 1 \cdot 8 = 8\)

### Step 4: Sum the Terms
Now sum these values:
\[ 27 + 270 + 180 + 8 = 485 \]

### Step 5: Multiply by 2
Since we have:
\[ (\sqrt{3} + \sqrt{2})^6 + (\sqrt{3} - \sqrt{2})^6 = 2 \cdot 485 = 970 \]

### Step 6: Analyze \((\sqrt{3} - \sqrt{2})^6\)
Next, we observe that \((\sqrt{3} - \sqrt{2})^6\) is a small positive number since \(\sqrt{3} - \sqrt{2} < 1\). Therefore:
\[ 0 < (\sqrt{3} - \sqrt{2})^6 < 1 \]

### Step 7: Establish Bounds
This implies:
\[ 970 - 1 < (\sqrt{3} + \sqrt{2})^6 < 970 \]
Thus:
\[ 969 < (\sqrt{3} + \sqrt{2})^6 < 970 \]

### Step 8: Conclusion
The smallest integer larger than \((\sqrt{3} + \sqrt{2})^6\) is:
\[ \boxed{970} \]