How do you factor #x^3-27v^3#?

Question

2974 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-10T00:19:49

#(x-3v)(x^2+3vx+9v^2)#

This problem is what is described as a Difference of Cubes

If we consider the equation:

#a^3-b^3# where #a# and #b# are perfect cubes, we can factorise the problem using the formula:

#(a – b)(a^2 + ab + b^2)#

In your case, if we consider the equation:

#x^3−27v^3#

We notice that not only are the #x# and #y# terms cubes, but 27 is also a cube #(3^3=27)#.

Therefore, if we substitute the values in to the Difference of Cubes formula, letting #x=a# and #3v=b# we get:

#(x-3v)(x^2+3vx+[3v]^2)#

Simplifying this we get the factored equation:

#(x-3v)(x^2+3vx+9v^2)#