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

  • Context: MHB 
  • Thread starter Thread starter Mathsonfire
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the factorization of the expression \(a^2(b^3-c^3)+b^2(c^3-a^3)+c^2(a^3-b^3\). Participants explore different approaches to factor this polynomial, engaging in technical reasoning and mathematical manipulation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests rearranging the expression to facilitate factorization, proposing an initial grouping of terms.
  • Another participant confirms that \((a-b)\) is a factor by substituting \(b=a\) and observing the result, which leads to the elimination of one potential answer.
  • There is a discussion about the cyclic symmetry of the expression, leading to the hypothesis that \((b-c)\) and \((c-a)\) are also factors, with suggestions to verify this through substitution.
  • Participants express appreciation for the methods used by others, indicating a collaborative atmosphere in exploring the factorization.
  • One participant points out a minor correction in the notation of the final factorization, noting a typographical error in the arrangement of terms.

Areas of Agreement / Disagreement

Participants generally agree on the presence of the factors \((a-b)\), \((b-c)\), and \((c-a)\), but the discussion includes various approaches and methods to arrive at the final factorization, indicating that multiple views and techniques are present without a definitive consensus on the best method.

Contextual Notes

Some participants rely on specific substitutions to test for factors, while others engage in rearranging terms, suggesting that the factorization process may depend on the approach taken. There are unresolved aspects regarding the clarity of the final expression and the implications of the corrections made during the discussion.

Mathsonfire
Messages
11
Reaction score
0
Pls hlp I am stuck.Pls give me a clue on this...
Question number 7
0e77c0e9f7b89c2ae89bdad75a8cbaa0.jpg
 
Mathematics news on Phys.org
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:

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

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

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

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

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

$$(a-b)(b-c)(c-a)(ab+bc+ca)$$
 
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.
 
@I love Serena that was a great method thanks...
@MarkFl thanks again...
 
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. ;)
 
I like Serena said:
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. :)
 
MarkFL said:
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.
 
I like Serena said:
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.
 
MarkFL said:
$$(a-b)(b-c)(c-a)(ab+bc+ca)$$
That "ca" should be "ac" :)
 
  • #10
Wilmer said:
That "ca" should be "ac" :)

I agree, however, I was arranging it the way it is shown in the attached photo. :)
 
  • #11
A thousand apologies of which you may have one!
 

Similar threads

MHB Find Solutions to a+b+c+d=4, a^2+b^2+c^2+d^2=6, a^3+b^3+c^3+d^3=94/9 in [0,2]
  • · Replies 2 ·
Replies
2
Views
1K
MHB Real Roots of Cubic Equation: $x^3+a^3x^2+b^3x+c^3=0$
  • · Replies 1 ·
Replies
1
Views
2K
MHB -c5.LCM and Prime Factorization of A,B,C
  • · Replies 3 ·
Replies
3
Views
1K
MHB What is the value of (a+b)^3+(b+c)^3+(c+a)^3 for the roots of 8x^3+1001x+2008=0?
  • · Replies 2 ·
Replies
2
Views
1K
MHB Proving $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$
  • · Replies 1 ·
Replies
1
Views
1K
High School Is the following fact significant? (the factorisation of A^3 +/- B^3)
  • · Replies 2 ·
Replies
2
Views
1K
MHB Solve for $d-b$: $a^5=b^4,\,c^3=d^2,\,c-a=19$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Find Product: $(a^4+b^4+c^4)$ and $(a^6+b^6+c^6)$
  • · Replies 3 ·
Replies
3
Views
1K
MHB Prove the inequality (a^3-c^3)/3≥abc((a-b)/c+(b-c)/a)
  • · Replies 3 ·
Replies
3
Views
2K
MHB Is the Product \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\) Divisible by 7?
  • · Replies 1 ·
Replies
1
Views
1K