How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$ How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?
It is part of a question about finding limits at mooculus.

A bit more roundabout way:
$x^4 - 1296 \ $ is a difference of two squares, as is $ \ x^2 - 36 \ $, so
$$(x^4 - 1296) \ = \ [ \ (x^2)^2 \ - \ (36^2) \ ] \ = \ (x^2 + 36) \ \cdot (x^2 - 36 ) $$
$$= \ (x^2 + 36) \cdot (x - 6) \cdot (x + 6) \ = \ (x^3 \ + \ 36x \ - \ 6x^2 - 216 ) \cdot (x + 6) \ . $$
EDIT: I missed that Πάρτη Κοηλί had done this going down to complex roots; this version "keeps things real"...
All of the answers I see at the present are fine -- there isn't any getting around that one needs to recognize that 1296 is $ \ 6^4 \ $ , or otherwise needs to find enough factors of 1296 to see a way to "break down" the polynomial.
