To factorize the expression $a^4 + a^2 + 1$, observe that it is a quadratic in terms of $a^2$. Let's rewrite it:

$a^4 + a^2 + 1 = (a^2)^2 + a^2 + 1$

This expression resembles the sum of a geometric series, which can be factorized using complex numbers. Specifically:

$a^4 + a^2 + 1 = \frac{a^6 - 1}{a^2 - 1}, \text{ provided } a \neq 1.$

However, in simpler real terms, $a^4 + a^2 + 1$ can also be factorized directly into:

$a^4 + a^2 + 1 = (a^2 + a + 1)(a^2 - a + 1)$

Verification of Factorization:

1. Multiply the factors $(a^2 + a + 1)(a^2 - a + 1)$: $(a^2 + a + 1)(a^2 - a + 1) = a^4 - a^2 + a^2 + a^2 - a + a - 1 = a^4 + a^2 + 1$

Final Answer:

$a^4 + a^2 + 1 = (a^2 + a + 1)(a^2 - a + 1)$

Let me know if you want further clarifications or have other questions.

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Expanding Understanding

1. How can you derive the formula for factoring such expressions systematically?
2. What is the geometric significance of $a^4 + a^2 + 1$ in complex roots of unity?
3. How does the factorization change if $a$ has constraints, such as being real or imaginary?
4. Can we generalize this factorization method for $a^{2n} + a^n + 1$?
5. How would this expression behave in modular arithmetic with certain primes?

Tip: Always verify your factorization by multiplying back to ensure correctness!