How do you factor #u^3 - 1#?

1 Answer
Sep 23, 2015

Answer:

#u^3-1 = (u-1)(u^2+u+1)#

Explanation:

There is a useful 'difference of cubes' identity:

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

In our example, #a = u# and #b = 1# :

#u^3-1 = u^3-1^3 = (u-1)(u^2+u*1+1^2) = (u-1)(u^2+u+1)#

In fact, in general we find:

#a^n - b^n = (a-b)(a^(n-1) + a^(n-2)b + a^(n-3)b^2 +...+ b^(n-1))#

When you multiply it out, most of the terms cancel.

When #n# is even, the second factor can be factorised further:

#a^(n-1) + a^(n-2)b + a^(n-3)b^2 +...+ b^(n-1)#

#= (a+b)(a^(n-2) + a^(n-4)b^2 + ... + b^(n-2))#

There are no more linear factors with Real coefficients, but there are quadratic factors with Real coefficients.

For example:

#a^6 - b^6 = (a-b)(a+b)(a^2-ab+b^2)(a^2+ab+b^2)#