Question 391773
how do you factor 27x to the third-8?

--------------------
That is a "difference of cubes".
---
Rewrite as (3x)^3 - 2^3
----------------------------
Factors:
(3x-2)((3x)^2+(3x)(2)+2^2)
---
(3x-2)(9x^2+6x+4)
=======================
Cheers,
Stan H.