Answer
$\frac{1}{8}(4x-1)(16x^2+4x+1)$
Work Step by Step
The formula for factoring the sum of two cubes is: $A^3+B^3=(A+B)(A^2-AB+B^2)$.
The formula for factoring the difference of two cubes is: $A^3-B^3=(A-B)(A^2+AB+B^2)$.
Hence here: $8x^4-\frac{x}{8}=\\=\frac{1}{8}(64x^3-1)\\=\frac{1}{8}((4x)^3-1^3)\\=\frac{1}{8}(4x-1)((4x)^2+1(4x)+1^2)\\=\frac{1}{8}(4x-1)(16x^2+4x+1)$