(abc^3 + ab^3c + a^3 bc) / (a^2 + b^2 + c^2 ) = abc

I don't understand how to show the steps to simplify this equation. Any help is well appreciated.

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- Sep 19th 2011, 09:43 AMgreengiltSimplifying the equation
(abc^3 + ab^3c + a^3 bc) / (a^2 + b^2 + c^2 ) = abc

I don't understand how to show the steps to simplify this equation. Any help is well appreciated. - Sep 19th 2011, 09:46 AMPlatoRe: Simplifying the equation
- Sep 19th 2011, 10:22 AMTheChazRe: Simplifying the equation
Another perspective, to help verify this identity (for at least one of a, b, c nonzero):

The numerator has a common factor "abc".

Factor this out, and you'll be left with... - Sep 19th 2011, 10:26 AMgreengiltRe: Simplifying the equation
I see the answer and what you are telling me works in my head like so:

$\displaystyle {ABC^3 + AB^3C + A^3BC } = ABC(A^2 + B^2 + C^2) $

However if, for example, I only had :

$\displaystyle {ABC^3 + AB^3C + A^3BC } \over {A^2 + B^2 + C^2} $

then when I try to simplify I get

$\displaystyle {ABC \cdot C^2 + ABC \cdot B^2 + ABC \cdot A^2} \over {A^2 + B^2 + C^2} $

and my answer ends up being

3ABC

Where am I going wrong? - Sep 19th 2011, 04:02 PMGSmithRe: Simplifying the equation
Like TheChaz said the numerator has a common factor "abc"

(abc^3 + ab^3c + a^3 bc) = abc(c^2 + b^2 + a^2) - Sep 19th 2011, 04:06 PMTheChazRe: Simplifying the equation
I'm guessing that you are going wrong in a terrible way.

Symbolically, you could (COULD, not SHOULD!) see the C^2 on top and likewise a C^2 on bottom, and literally just erase them from the page. Same with A^2 and B^2.

In that case, you would have ABC(eraser mark) + ABC(eraser mark) + ABC(eraser mark), all divided by (eraser mark + eraser mark + eraser mark)

And maybe you're thinking that since you have ABC three times, that this is 3ABC.

This is terribly wrong.

"Cancel factors, not terms" is a concise way to avoid this problem. I'll let someone else fill in the distinction.