(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.
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?
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.