Sure, let's factorize the expression [tex]\(a^4 + a^2 + 1\)[/tex] step-by-step.
First, observe that the expression can be rewritten in a form that helps us recognize patterns:
[tex]\[a^4 + a^2 + 1 = a^4 + a^2 + 1^2\][/tex]
Interestingly, this form hints at a relationship similar to the sum of cubes or the properties of certain polynomial identities. Let's attempt a factorization by recognizing that [tex]\(a^2 = x\)[/tex], where [tex]\(x = a^2\)[/tex]:
[tex]\[a^4 + a^2 + 1 \equiv x^2 + x + 1\][/tex]
We need to factor [tex]\(x^2 + x + 1\)[/tex]. With some algebraic insight, we recognize this polynomial as a known quadratic that factors into:
[tex]\[x^2 + x + 1 = (x - \omega)(x - \omega^2)\][/tex]
where [tex]\( \omega \)[/tex] and [tex]\( \omega^2 \)[/tex] are the non-real cube roots of unity. However, expressing this back in terms of [tex]\(a\)[/tex], we get:
[tex]\[a^4 + a^2 + 1 = (a^2 - a + 1)(a^2 + a + 1)\][/tex]
Therefore, the factorized form of [tex]\(a^4 + a^2 + 1\)[/tex] is:
[tex]\[(a^2 - a + 1)(a^2 + a + 1)\][/tex]
This is the desired factorization.
