Welcome to our community

Be a part of something great, join today!

Problem Of The Week #485 October 4th 2021

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,937
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,937
Congratulations to the following members for their correct solution!(Cool)

1. kaliprasad
2. lfdahl

Solution from lfdahl :
Given $P(r) = r^3-2r+1 = 0$:

\[0 = [P(r)]^2 = \left [ r^3-2r+1 \right ]^2 \\\\ = r^6-4r^4+2r^3+4r^2-4r+1 \\\\ = \underbrace{r^6 - 4r^4 + 4r^2-1}_{= Q(r^2)} +\underbrace{2r^3-4r + 2}_{= 2P(r)} = Q(r^2)\;\;\;\; q.e.d.\]
 
Status
Not open for further replies.