
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
May 27th, 2014, 08:29 PM  #1 
Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 227 Thanks: 30 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology  A factorization
$x^3 + y^3 + z^3  3xyz$ This was the hardest problem in my 1st grade high school book. I couldn't solve it for the life of me. Maybe it will be pretty easy for you guys. 
May 28th, 2014, 03:01 AM  #2 
Newbie Joined: May 2014 From: Pune Posts: 12 Thanks: 0 
Well, I didn't get your question. Can you please clarify it?

May 28th, 2014, 03:09 AM  #3 
Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 227 Thanks: 30 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology 
Show the expression as a product of expressions. As in $x^2y^2 = ($xy$)($x+y$). 
May 28th, 2014, 06:25 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,993 Thanks: 652 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Apparently the result is $\displaystyle x^3 + y^3 + z^3  3xyz \equiv (x+y+x)(x^2 + y^2 + z^2  xy  xz  yz)$ according to the web. It is an identity. I must admit that I'd never seen it before until today. 
May 28th, 2014, 06:54 AM  #5 
Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 227 Thanks: 30 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology 
You wrote it as a congruence. Shouldn't it be =? Anyway could you prove it, please? I mean, it's easy to multiply, bur I'd like as if you had the left term and try to figure out a way to factorize it.

May 29th, 2014, 02:31 AM  #6  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,993 Thanks: 652 Math Focus: Physics, mathematical modelling, numerical and computational solutions  No, because the relation is valid for all choices of x, y and z. It is an identity, not an equation. Quote:
i) pull out x + y + z from each cube, subtracting the other two terms ii) rearrange the extra subtracted terms for content involving xy, xz or yz and give each part an xyz from the 3xyz term: iii) Factorise from each large bracket xy, xz or yz iv) Factorise (x+y+z) from every term Prove $\displaystyle x^3 + y^3 + z^3 3xyz \equiv (x+y+z)( x^2 + y^2 + z^2  xy  xz  yz)$ LHS $\displaystyle = x^3 + y^3 + z^3 3xyz$ $\displaystyle = x^2(x+y+z)  x^2y  x^2z + y^2(x+y+z)  y^2x  y^2z + z^2(x+y+z)  z^2x  z^2y  3xyz$ $\displaystyle = x^2(x+y+z) + y^2(x+y+z) + z^2(x+y+z)  (x^2y + y^2x +xyz)  (x^2z + z^2x + xyz)  (y^2z + z^2y + xyz)$ $\displaystyle = x^2(x+y+z) + y^2(x+y+z) + z^2(x+y+z)  xy(x+y+z)  xz(x+y+z)  yz(x+y+z)$ $\displaystyle = (x+y+z)( x^2 + y^2 + z^2  xy  xz  yz) = $ RHS  
May 29th, 2014, 04:12 AM  #7 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 466 Thanks: 40 
May I ask what this identity is used for? Congratulations on the factorisation. 
May 29th, 2014, 04:31 AM  #8 
Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 227 Thanks: 30 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology 
I don't know if it's used for anything. It's just a problem from my high school book.

May 29th, 2014, 04:48 AM  #9 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,993 Thanks: 652 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
A quick browse on the internet suggests that it can be used to factorise specific cubic equations (although I prefer the factor theorem and long division myself). For example: Factorise $\displaystyle x^3  6x + 9$ This is equal to $\displaystyle x^3 + y^3 + z^3 3xyz$ where $\displaystyle y=1, z=2$ Therefore: $\displaystyle x^3  6x + 9 = (x + 1 + 2)(x^2 + 1^2 + 2^2  x  2x  2)$ $\displaystyle = (x+3)(x^23x+3)$ I gather it is more useful if you have two unknowns, so you can factorise stuff like: $\displaystyle x^3 + y^3  6xy + 8$ 
June 3rd, 2014, 11:09 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 17,221 Thanks: 1294 
$x^3 + y^3 + z^3  3xyz \equiv (x + y + z)(x + \omega y + \omega^2z)(x + \omega^2y + \omega z),$ where $\omega$ is a complex cube root of 1.


Tags 
factorization 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Factorization  Wissam  Number Theory  39  December 2nd, 2010 12:29 PM 
Factorization help  advancedfunctions  Calculus  3  March 9th, 2010 07:31 PM 
Factorization  abhijith  Algebra  1  November 7th, 2009 10:23 AM 
Factorization  johnny  Algebra  2  September 6th, 2009 08:35 PM 
Factorization Help  kirankai  Elementary Math  2  February 25th, 2009 12:46 AM 