1. ## 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.

2. ## Re: Simplifying the equation

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 $\frac{A}{B}=C$ then it must be $A=BC~.$

3. ## Re: 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...

4. ## Re: Simplifying the equation

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

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

However if, for example, I only had :

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

then when I try to simplify I get

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

and my answer ends up being

3ABC

Where am I going wrong?

5. ## Re: 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)

6. ## Re: 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.