Thread: Simplifying 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.

Originally Posted by greengilt

(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.

Hint: if then it must be

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...

I see the answer and what you are telling me works in my head like so:



However if, for example, I only had :



then when I try to simplify I get



and my answer ends up being

3ABC

Where am I going wrong?

Like TheChaz said the numerator has a common factor "abc"

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

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.