## Precalculus (6th Edition) Blitzer

$(3x-1)(3x+1)(9x^2+1)$
The given expression can be written as $(9x^2)^2-1^2.$ The expression above is a difference of two squares. RECALL: $a^2-b^2=(a-b)(a+b).$ Factor the difference of two squares using the formula above with $a=9x^2$ and $b=1$ to obtain $(9x^2-1)(9x^2+1) \\=[(3x)^2-1^2](9x^2+1).$ Factor the difference of two squares with $a=3x$ and $b=1$ to obtain $(3x-1)(3x+1)(9x^2+1).$