perplexus.info :: Just Math : Quick fatorization

Quick fatorization (Posted on 2024-01-06)

Factorize (a+2b-3c)³+(b+2c-3a)³+(c+2a-3b)³

No Solution Yet Submitted by Danish Ahmed Khan

No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)

Puzzle Solution

Comment 1 of 1

Let X= a+2b+3c, Y= b+2c-3a, and: Z=c+2a-3b

Then, we observe that:

X+Y+Z = 0

Or, Y+Z=-X

Therefore, X^3+Y^3+Z^3

= Y^3+Z^3-(Y+Z)^3

=-3YZ(Y+Z)

= -3(b+2c-3a)(c+2a-3b)(3c-a-2b)

Similarly, substituting Z+X=-Y, and: X+Y=-Z, we would obtain:

-3(c+2a-3b)(a+2b-3c)(3a-b-2c), and: -3(b+2c-3a)(c+2a-3b)(3c-a-2b) as two other valid factorisations.

---------------------------------------------------------------

IMHO: The title should have been "Quick Factorisation"

Edited on January 6, 2024, 11:47 am

Posted by K Sengupta on 2024-01-06 11:41:01

Please log in:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (15)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info