Question 161000
{{{216p^3 - 1=(6p)^3-1^3=(6p-1)((6p)^2+(6p)(1)+1^2)=(6p-1)(36p^2+6p+1)}}}



So {{{216p^3 - 1}}} completely factors to {{{(6p-1)(36p^2+6p+1)}}}



Note: use the difference of cubes formula to factor. the difference of cubes formula is {{{A^3-B^3=(A+B)(A^2+AB+B^2)}}}