
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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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: 1,963 Thanks: 639 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: 1,963 Thanks: 639 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: 1,963 Thanks: 639 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: 16,932 Thanks: 1253 
$x^3 + y^3 + z^3  3xyz \equiv (x + y + z)(x + \omega y + \omega^2z)(x + \omega^2y + \omega z),$ where $\omega$ is a complex cube root of 1.


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