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


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May 27th, 2014, 08:29 PM   #1
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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.
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May 28th, 2014, 03:01 AM   #2
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Well, I didn't get your question. Can you please clarify it?
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May 28th, 2014, 03:09 AM   #3
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Show the expression as a product of expressions. As in

$x^2-y^2 = ($x-y$)($x+y$).
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May 28th, 2014, 06:25 AM   #4
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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.
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May 28th, 2014, 06:54 AM   #5
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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.
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May 29th, 2014, 02:31 AM   #6
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Quote:
Originally Posted by raul21 View Post
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 View Post
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
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May 29th, 2014, 04:12 AM   #7
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May I ask what this identity is used for?

Congratulations on the factorisation.
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May 29th, 2014, 04:31 AM   #8
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I don't know if it's used for anything. It's just a problem from my high school book.
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May 29th, 2014, 04:48 AM   #9
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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:

$\displaystyle x^3 + y^3 - 6xy + 8$

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June 3rd, 2014, 11:09 AM   #10
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$x^3 + y^3 + z^3 - 3xyz \equiv (x + y + z)(x + \omega y + \omega^2z)(x + \omega^2y + \omega z),$ where $\omega$ is a complex cube root of 1.
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