Factorisation of a^2(b^3−c^3)+b^2(c^3−a^3)+c^2(a^3−b^3)

[Mathsonfire]

Pls hlp I am stuck.Pls give me a clue on this...
Question number 7

 

[MarkFL]

We are given to factor:

$$a^2(b^3-c^3)+b^2(c^3-a^3)+c^2(a^3-b^3)$$

I think I would arrange as:

$$c^2(a^3-b^3)-(a^3b^2-a^2b^3)-c^3(a^2-b^2)$$

Factor terms:

$$c^2(a-b)(a^2+ab+b^2)-a^2b^2(a-b)-c^3(a+b)(a-b)$$

Factor out factor common to all 3 terms:

$$(a-b)(c^2(a^2+ab+b^2)-a^2b^2-c^3(a+b))$$

In the second factor, expand and group on like $a$ terms:

$$(a-b)((c^2-b^2)a^2+(bc^2-c^3)a+(b^2c^2-bc^3))$$

$$(a-b)(-(c+b)(b-c)a^2+ac^2(b-c)+bc^2(b-c))$$

$$(a-b)(b-c)(-(c+b)a^2+ac^2+bc^2)$$

$$(a-b)(b-c)((c-a)ac+(a+c)(c-a)b)$$

$$(a-b)(b-c)(c-a)(ac+(a+c)b)$$

$$(a-b)(b-c)(c-a)(ab+bc+ca)$$

 

[I like Serena]

Let me try to give a bit of rationale to MarkFL's approach.

Two of the given answers contain (a-b), so let's try to confirm if (a-b) is a factor or not.
If (a-b) is a factor, then substituting b=a must yield 0... and it does (check it).
Consequently we can already eliminate answer (2).

So we try to extract (a-b). We can extract it from $c^2(a^3-b^3)$ directly, so it makes sense to put that term at the front.
We cannot extract it directly from the first and second terms, so we need to rearrange those so that we can extract (a-b), which is what Mark did.

From cyclic symmetry we can tell that (b-c) and (c-a) will also be factors, or otherwise we can check in the same way by substituting c=b respectively a=c.
Extract those factors as well in the same fashion, which brings us to Mark's final result.

 

[Mathsonfire]

@I love Serena that was a great method thanks...
@MarkFl thanks again...

 

[I like Serena]

Yep. The @ symbol triggered the spam detector, since it looks like an email address posted by a new user.
Let me clean that up.

Edit: Too late. Mark already did. ;)

 

[MarkFL]

Yep. The @ symbol triggered the spam detector, since it looks like an email address posted by a new user.
Let me clean that up.

Edit: Too late. Mark already did. ;)

A lot of sites now have forum software that allows tagging users with the @ symbol, so I may have to rethink using that as a trigger. :)

 

[I like Serena]

A lot of sites now have forum software that allows tagging users with the @ symbol, so I may have to rethink using that as a trigger. :)

Ooh, it would be really cool if typing @ would trigger user name markup somehow (instead of triggering the spam filter). Ah well, that is probably asking for too much.

 

[MarkFL]

Ooh, it would be really cool if typing @ would trigger user name markup somehow (instead of triggering the spam filter). Ah well, that is probably asking for too much.

DBTech offers an addon that allows true user tagging using the @ symbol, which I've tested some time ago out of curiosity on my dev site. On XF and IPS sites, typing the @ symbol and one ore more letters will even bring up a popup menu of usernames fitting the pattern typed.

Then the user tagged receives a notification that someone has tagged them, and a link to the post is provided.

 

[Wilmer]

$$(a-b)(b-c)(c-a)(ab+bc+ca)$$

That "ca" should be "ac" :)

 

[MarkFL]

That "ca" should be "ac" :)

I agree, however, I was arranging it the way it is shown in the attached photo. :)

 

[Wilmer]

A thousand apologies of which you may have one!