How do you factor #27w^3z-z^4w^6#?

1 Answer
Apr 7, 2017

Answer:

#27w^3z-z^4w^6 = w^3z(3-zw)(9+3zw+z^2w^2)#

Explanation:

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

We will use this with #a=3# and #b=zw#.

#color(white)()#
Given:

#27w^3z-z^4w^6#

First note that both of the terms are divisible by #w^3# and by #z#, so by #w^3z#. So we can separate that out as a factor first:

#27w^3z-z^4w^6 = w^3z(27-z^3w^3)#

#color(white)(27w^3z-z^4w^6) = w^3z(3^3-(zw)^3)#

#color(white)(27w^3z-z^4w^6) = w^3z(3-zw)(3^2+3zw+(zw)^2)#

#color(white)(27w^3z-z^4w^6) = w^3z(3-zw)(9+3zw+z^2w^2)#

This is as far as we can go with Real coefficients.

The remaining quartic factor #(9+3zw+z^2w^2)# can be factored further, but only with the use of Complex coefficients.