1. **State the problem:** Factorize the expression $x^4 + x^2 y^2 + y^4$. 2. **Recall the formula:** This expression resembles a quadratic in terms of $x^2$ and $y^2$. We can try to factor it as a product of two quadratic expressions. 3. **Rewrite the expression:** Let $a = x^2$ and $b = y^2$. Then the expression becomes $a^2 + ab + b^2$. 4. **Check for factorization:** The expression $a^2 + ab + b^2$ does not factor over the real numbers into linear factors, but it can be factored over complex numbers or expressed using sum of cubes formula or as a product of quadratics with complex coefficients. 5. **Use the identity:** Note that $a^2 + ab + b^2 = (a + \omega b)(a + \omega^2 b)$ where $\omega$ is a complex cube root of unity satisfying $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. 6. **Substitute back:** So, $$ x^4 + x^2 y^2 + y^4 = (x^2 + \omega y^2)(x^2 + \omega^2 y^2) $$ where $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. 7. **Summary:** The polynomial is irreducible over the reals but factorizes over the complex numbers as above. **Final answer:** $$ x^4 + x^2 y^2 + y^4 = (x^2 + \omega y^2)(x^2 + \omega^2 y^2) $$