Answer :

To factorize x^{4} + x^{2} + 1, consider the expression (x^{2} + 1)^{2}.

We have the identity (a + b)^{2} = a^{2} + 2ab + b^{2}

⇒ (x^{2} + 1)^{2} = x^{4} + 2x^{2} + 1

Rearranging terms in the equation by moving one x^{2} term to the left hand side, we have,

(x^{2} + 1)^{2} – x^{2} = x^{4} + x^{2} + 1

⇒ x^{4} + x^{2} + 1 = (x^{2} + 1)^{2} – x^{2}

Using the identity (a + b)(a – b) = a^{2} – b^{2}, we have,

x^{4} + x^{2} + 1 = [(x^{2} + 1) + x][(x^{2} + 1) – x]

∴ x^{4} + x^{2} + 1 = (x^{2} + x + 1) (x^{2} – x + 1)

Hence, the factors of x^{4} + x^{2} + 1 are (x^{2} + x + 1) and (x^{2} – x + 1).

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