Factorise `x ^(8) + x ^(4) +1`

Answer

To factorize the expression \( x^8 + x^4 + 1 \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ x^8 + x^4 + 1 \]
Notice that \( x^8 \) can be rewritten as \( (x^4)^2 \). This gives us:
\[ (x^4)^2 + x^4 + 1 \]

### Step 2: Use a Substitution
Let \( y = x^4 \). Then, the expression becomes:
\[ y^2 + y + 1 \]

### Step 3: Factor the Quadratic Expression
The expression \( y^2 + y + 1 \) can be factored using the formula for the sum of cubes. We recognize that:
\[ y^2 + y + 1 = \frac{y^3 - 1}{y - 1} \quad \text{(for } y \neq 1\text{)} \]
This can be factored as:
\[ y^2 + y + 1 = (y - \omega)(y - \omega^2) \]
where \( \omega = e^{2\pi i / 3} \) is a primitive cube root of unity.

### Step 4: Substitute Back
Now we substitute back \( y = x^4 \):
\[ (x^4 - \omega)(x^4 - \omega^2) \]

### Step 5: Final Expression
Thus, the factorization of the original expression \( x^8 + x^4 + 1 \) is:
\[ (x^4 - \omega)(x^4 - \omega^2) \]