Find the value of `(1+ i)^(6) + (1-i)^(6)`

Options

A. `16i`
B. `0`
C. `-16i`
D. `1`

Explanation

To solve the expression \((1 + i)^{6} + (1 - i)^{6}\), we can follow these steps:

### Step 1: Rewrite the expression
We can express the terms in the expression as:
\[ (1 + i)^{6} + (1 - i)^{6} \]

### Step 2: Use the binomial theorem
Using the binomial theorem, we can expand both terms:
\[ (1 + i)^{6} = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (i)^{k} \]
\[ (1 - i)^{6} = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (-i)^{k} \]

### Step 3: Combine the expansions
Now, we combine the two expansions:
\[ (1 + i)^{6} + (1 - i)^{6} = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (i)^{k} + \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (-i)^{k} \]

### Step 4: Analyze the terms
Notice that for even \(k\), \((i)^{k} + (-i)^{k} = 2(i)^{k}\) and for odd \(k\), \((i)^{k} + (-i)^{k} = 0\). Thus, we only need to consider the even \(k\) terms:
\[ = 2 \sum_{k \text{ even}} \binom{6}{k} (1)^{6-k} (i)^{k} \]

### Step 5: Calculate the even terms
The even values of \(k\) from 0 to 6 are 0, 2, 4, and 6. Thus, we have:
\[ = 2 \left( \binom{6}{0} (1)^{6} (i)^{0} + \binom{6}{2} (1)^{4} (i)^{2} + \binom{6}{4} (1)^{2} (i)^{4} + \binom{6}{6} (1)^{0} (i)^{6} \right) \]

Calculating each term:
- For \(k=0\): \(\binom{6}{0} (1)^6 (i)^0 = 1\)
- For \(k=2\): \(\binom{6}{2} (1)^4 (i)^2 = 15(-1) = -15\)
- For \(k=4\): \(\binom{6}{4} (1)^2 (i)^4 = 15(1) = 15\)
- For \(k=6\): \(\binom{6}{6} (1)^0 (i)^6 = 1(-1) = -1\)

### Step 6: Sum the contributions
Now, summing these contributions:
\[ = 2 \left( 1 - 15 + 15 - 1 \right) = 2 \cdot 0 = 0 \]

### Final Answer
Thus, the value of \((1 + i)^{6} + (1 - i)^{6}\) is:
\[ \boxed{0} \]