How do you factor #a^3b^6-b^3#?

Question

2481 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-16T16:58:01

#b^3(a^3b^3 - b^3)#

#a^3b^6 - b^3#

#b^6 = b^3 cdot b^3#

#(a^3b^3b^3 - b^3)#

factorizing;

#b^3(a^3b^3 - b^3)#

Answer 2 · 2018-05-16T17:00:16

#a^3b^6-b^3=b^3(ab-1)(a^2b^2+ab+1)#

We want to simplify #a^3b^6-b^3#. The first thing to do is to factor out #b^3# to give #b^3(a^3b^3-1)#.

If we look carefully, we can see that #b^3(a^3b^3-1)=b^3((ab)^3-1^3)#. So now we can use a formula to simplify even more:

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

So

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