My Math Forum A factorization

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 May 27th, 2014, 08:29 PM #1 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 233 Thanks: 31 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: 233 Thanks: 31 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 $x^2-y^2 = ($x-y$)($x+y$).  May 28th, 2014, 06:25 AM #4 Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,039 Thanks: 674 Math Focus: Physics, mathematical modelling, numerical and computational solutions Apparently the result is$\displaystyle x^3 + y^3 + z^3 - 3xyz \equiv (x+y+x)(x^2 + y^2 + z^2 - xy - xz - yz)$according to the web. It is an identity. I must admit that I'd never seen it before until today. Thanks from raul21  May 28th, 2014, 06:54 AM #5 Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 233 Thanks: 31 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology You wrote it as a congruence. Shouldn't it be =? Anyway could you prove it, please? I mean, it's easy to multiply, bur I'd like as if you had the left term and try to figure out a way to factorize it. May 29th, 2014, 02:31 AM #6 Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,039 Thanks: 674 Math Focus: Physics, mathematical modelling, numerical and computational solutions Quote:  Originally Posted by raul21 You wrote it as a congruence. Shouldn't it be =? No, because the relation is valid for all choices of x, y and z. It is an identity, not an equation. Quote:  Originally Posted by raul21 Anyway could you prove it, please? I mean, it's easy to multiply, bur I'd like as if you had the left term and try to figure out a way to factorize it. Just figured out a sequence of factorisation to prove the identity (with the benefit of hindsight)... i) pull out x + y + z from each cube, subtracting the other two terms ii) rearrange the extra subtracted terms for content involving xy, xz or yz and give each part an xyz from the 3xyz term: iii) Factorise from each large bracket xy, xz or yz iv) Factorise (x+y+z) from every term Prove$\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) = $RHS  May 29th, 2014, 04:12 AM #7 Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 May I ask what this identity is used for? Congratulations on the factorisation.  May 29th, 2014, 04:31 AM #8 Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 233 Thanks: 31 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology I don't know if it's used for anything. It's just a problem from my high school book.  May 29th, 2014, 04:48 AM #9 Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,039 Thanks: 674 Math Focus: Physics, mathematical modelling, numerical and computational solutions A quick browse on the internet suggests that it can be used to factorise specific cubic equations (although I prefer the factor theorem and long division myself). For example: Factorise$\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^2-3x+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$Thanks from raul21  June 3rd, 2014, 11:09 AM #10 Global Moderator Joined: Dec 2006 Posts: 17,740 Thanks: 1361$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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