How do you factor #x^3-729 #?

1 Answer
Dec 10, 2015

Answer:

# (x-9)(x^2+9x+81)#

Explanation:

You have to observe that #729=9^3#, and so #x^3-729# is a difference of cubes.

In general, you have that

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

and #(a^2+ab+b^2)# is no longer factorizable (it is also known as "false square", because it resembles #(a+b)^2#, but the mixed product is halved).

In your case, #a=x# and #b=9#, so we have

#x^3-9^3 = (x-9)(x^2+9x+81)#