Evaluate: (1 + i)^6 + (1 - i)^3

We are tasked with evaluating the expression:

(1 + i)^6 + (1 - i)^3

Step 1: Simplify (1 + i)^6
First, we'll simplify (1 + i)^6. We use the binomial expansion formula for this:

(a + b)^n = Σ (nCk) * a^(n-k) * b^k

For (1 + i)^6, we expand using the binomial theorem:

(1 + i)^6 = Σ (6Ck) * 1^(6-k) * i^k for k = 0 to 6

Calculating the terms:

For k = 0: (6C0) * 1^6 * i^0 = 1 * 1 * 1 = 1
For k = 1: (6C1) * 1^5 * i^1 = 6 * 1 * i = 6i
For k = 2: (6C2) * 1^4 * i^2 = 15 * 1 * (-1) = -15
For k = 3: (6C3) * 1^3 * i^3 = 20 * 1 * (-i) = -20i
For k = 4: (6C4) * 1^2 * i^4 = 15 * 1 * 1 = 15
For k = 5: (6C5) * 1^1 * i^5 = 6 * 1 * i = 6i
For k = 6: (6C6) * 1^0 * i^6 = 1 * 1 * (-1) = -1
Now, adding all the terms together:

(1 + i)^6 = 1 + 6i - 15 - 20i + 15 + 6i - 1

Simplify:

= (1 - 15 + 15 - 1) + (6i - 20i + 6i)

= 0 - 8i

Thus, (1 + i)^6 = -8i.

Step 2: Simplify (1 - i)^3
Next, we simplify (1 - i)^3. Using the binomial expansion again:

(1 - i)^3 = Σ (3Ck) * 1^(3-k) * (-i)^k for k = 0 to 3

Calculating the terms:

For k = 0: (3C0) * 1^3 * (-i)^0 = 1 * 1 * 1 = 1
For k = 1: (3C1) * 1^2 * (-i)^1 = 3 * 1 * (-i) = -3i
For k = 2: (3C2) * 1^1 * (-i)^2 = 3 * 1 * (-1) = -3
For k = 3: (3C3) * 1^0 * (-i)^3 = 1 * 1 * i = i
Now, adding all the terms together:

(1 - i)^3 = 1 - 3i - 3 + i

Simplify:

= (1 - 3) + (-3i + i)

= -2 - 2i

Thus, (1 - i)^3 = -2 - 2i.

Step 3: Add the results
Now, we add the two results together:

(1 + i)^6 + (1 - i)^3 = (-8i) + (-2 - 2i)

Simplify:

= -2 - 8i - 2i

= -2 - 10i

Thus, the final answer is:

-2 - 10i