Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)

Printable View

- Oct 1st 2009, 04:03 AManshulbshahA factorisation question
Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)

- Oct 1st 2009, 06:52 AMSoroban
Hello, anshulbshah!

I expanded it, then repeatedly factored "by grouping".

If there is a more direct way, I hope someone finds it.

Quote:

Factor: .

. .

. .

. .

. .

. .

. .

. .

. .

. .

. .

. .

. .

. .

Answer: .

- Oct 1st 2009, 07:13 AMmasters
- Oct 1st 2009, 07:55 AMGrandad
Hello anshulbshahHaving seen Soroban's answer, I suppose this is being wise after the event, but here's a quicker method, if you spot it!

If we put in the expression, we get:

so is a factor.

Similarly and are also factors.

Since the expression is of degree in , the remaining factor is linear, and by symmetry is therefore for some constant .

So

Compare coefficients of, say, (taking the from the first factor, the from the second, the from the third and the from the fourth)

Or did I cheat a bit?

Grandad

- Oct 1st 2009, 09:37 AMpacman
great work soroban and grandad, whew!

http://www.mathhelpforum.com/math-he...5ea190ad-1.gif = -(a - b)(a - c)(b - c)(a + b + c).

as was observed by grandad, a = b = c. But if you let a = -(b + c); the expression vanishes too.

Then from a = -(b + c), we have (a + b + c) as a factor also.