We are tasked with factoring the expression $x^{12}y^{18} + 1$. To solve this, notice that it is a sum of cubes because:

$x^{12}y^{18} + 1 = (x^4y^6)^3 + 1^3$

We can apply the sum of cubes factorization formula:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Here, $a = x^4y^6$ and $b = 1$, so applying the formula:

$x^{12}y^{18} + 1 = (x^4y^6 + 1)\left((x^4y^6)^2 - x^4y^6 \cdot 1 + 1^2\right)$

Simplifying the second factor:

$(x^4y^6)^2 = x^8y^{12}, \quad x^4y^6 \cdot 1 = x^4y^6, \quad 1^2 = 1$

Thus, the expression becomes:

$x^{12}y^{18} + 1 = (x^4y^6 + 1)(x^8y^{12} - x^4y^6 + 1)$

This matches the second option in the image.

Would you like more details or have any questions about this? Here are five related questions to deepen your understanding:

1. How does the difference of cubes formula differ from the sum of cubes?
2. What are other examples of using the sum of cubes formula in algebra?
3. How can we identify whether an expression is factorable using cube identities?
4. What happens if we change the powers in the original expression to different values?
5. Can this factorization method be extended to more complex polynomials?

Tip: Always look for common factorization patterns like sum or difference of cubes when dealing with higher powers!