1. The problem is to factor the expression $81x^{12} - 1$. 2. Recognize this as a difference of squares: $a^2 - b^2 = (a - b)(a + b)$. 3. Here, $81x^{12} = (9x^6)^2$ and $1 = 1^2$, so: $$81x^{12} - 1 = (9x^6)^2 - 1^2 = (9x^6 - 1)(9x^6 + 1)$$ 4. Next, factor $9x^6 - 1$ again as a difference of squares: $$9x^6 - 1 = (3x^3)^2 - 1^2 = (3x^3 - 1)(3x^3 + 1)$$ 5. The term $9x^6 + 1$ is a sum of squares and cannot be factored further over the reals. 6. Therefore, the full factorization is: $$81x^{12} - 1 = (3x^3 - 1)(3x^3 + 1)(9x^6 + 1)$$ 7. This matches the given factorization. Final answer: $$(3x^{3} - 1)(3x^{3} + 1)(9x^{6} + 1)$$