Prove that: `x^4=4=(x+1+i)(x+1-i)(x-1+i(x-1-i)dot)`

Answer

To prove that \( x^4 + 4 = (x + 1 + i)(x + 1 - i)(x - 1 + i)(x - 1 - i) \), we will start by expanding the right-hand side (RHS) and simplifying it step by step.

### Step 1: Group the terms
The RHS can be grouped as follows:
\[ (x + 1 + i)(x + 1 - i) \quad \text{and} \quad (x - 1 + i)(x - 1 - i) \]

### Step 2: Expand the first group
Using the difference of squares formula \( (a + b)(a - b) = a^2 - b^2 \):
\[ (x + 1 + i)(x + 1 - i) = (x + 1)^2 - i^2 \]
Calculating \( (x + 1)^2 \):
\[ (x + 1)^2 = x^2 + 2x + 1 \]
Since \( i^2 = -1 \), we have:
\[ -i^2 = -(-1) = 1 \]
Thus,
\[ (x + 1 + i)(x + 1 - i) = (x^2 + 2x + 1) + 1 = x^2 + 2x + 2 \]

### Step 3: Expand the second group
Similarly, for the second group:
\[ (x - 1 + i)(x - 1 - i) = (x - 1)^2 - i^2 \]
Calculating \( (x - 1)^2 \):
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Again, since \( i^2 = -1 \):
\[ -i^2 = 1 \]
Thus,
\[ (x - 1 + i)(x - 1 - i) = (x^2 - 2x + 1) + 1 = x^2 - 2x + 2 \]

### Step 4: Combine the results
Now we can combine the results from both groups:
\[ (x^2 + 2x + 2)(x^2 - 2x + 2) \]

### Step 5: Expand the combined expression
Using the distributive property (FOIL method):
\[ = x^2(x^2 - 2x + 2) + 2x(x^2 - 2x + 2) + 2(x^2 - 2x + 2) \]
Calculating each term:
1. \( x^2 \cdot x^2 = x^4 \)
2. \( x^2 \cdot (-2x) = -2x^3 \)
3. \( x^2 \cdot 2 = 2x^2 \)
4. \( 2x \cdot x^2 = 2x^3 \)
5. \( 2x \cdot (-2x) = -4x^2 \)
6. \( 2x \cdot 2 = 4x \)
7. \( 2 \cdot x^2 = 2x^2 \)
8. \( 2 \cdot (-2x) = -4x \)
9. \( 2 \cdot 2 = 4 \)

Now, combine all these terms:
\[ x^4 + (-2x^3 + 2x^3) + (2x^2 - 4x^2 + 2x^2) + (4x - 4x) + 4 \]
This simplifies to:
\[ x^4 + 0 + 0 + 0 + 4 = x^4 + 4 \]

### Conclusion
Thus, we have shown that:
\[ x^4 + 4 = (x + 1 + i)(x + 1 - i)(x - 1 + i)(x - 1 - i) \]
This completes the proof.