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IIB Lecture 1 Example 1.2(a)

 
 
FanJimmy的头像
IIB Lecture 1 Example 1.2(a)
FanJimmy - 2024年05月14日 Tuesday 15:18
 

I don't really get the part where the teacher said that we could just make (a+b)^3-3ab(a+b+c)+c^3 into (a+b+c)*[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c). Could somebody explain please?

 
LensmireJohn的头像
Re: IIB Lecture 1 Example 1.2(a)
LensmireJohn - 2024年05月14日 Tuesday 16:23
 

First off, let me just mention that probably the easiest way to prove the identity:$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - ac - bc)$$is to just expand everything on the right-hand side and find that it matches up.

That being said, it can be good to practice with some other identities and prove the formula as done in the video lecture. When you get to $(a+b)^3-3ab(a+b+c)+c^3$ note that $(a+b)^3 + c^3$ is the sum of two cubes. So this portion can be factored using the sum of cubes formula, which turns it into  $(a+b+c)*[(a+b)^2-c(a+b)+c^2]$. Hope this helps!

FanJimmy的头像
Re: IIB Lecture 1 Example 1.2(a)
FanJimmy - 2024年05月14日 Tuesday 19:02
 

Oh, Thank you! I get it now. Originally, when I saw this problem, I just proved it by expanding. Thanks!