Question

r^6 - 3r^4 + 3r^2 - 1 = 0 .. How to solve this differential equation ???

Answer 1

Factorise it if you want.

Though it may be easier to let x = r^2 than factorise that (it's a standard expansion if you recognise it; if not you can find a factor (or three) by inspection). Then
\[r = \pm \sqrt{x} \]

Answer 2

ohh yea i get it! thanks alot !

Answer 3

does this help r=1 multplicity 3
r=-1 multplicity 3

Answer 4

:)
If you're interesting, it actually factorises it it's original for as:
\[(r-1)^3(r+1)^3 \]

Which I think is quite quaint.

Answer 5

i tried to factor don't look at that first part where i factor r^4 of the first two terms

Answer 6

i got lucky and synthetic division worked since all r all are rational zeros

Answer 7

@myininaya. thanks it sure did help

Answer 8

@newton. but u will have to factorise to get that answer, no ? we cant figure it out directly, can we ?

Answer 9

\[(r-1)^3(r+1)^3 = 0 \iff r = \pm 1 \]

Answer 10

yes yes i get it. what i meant to say was that we will have to factorise the original expression using division to reach to this answer ?

Answer 11

Not necessarily using division, you could compare coefficients of something.

Answer 12

(after spotting a few factors by inspection - in this case they are so obvious, but normally trying -2, -1, 1, 2 will get at least one)

Answer 13

look i have something else if you didnt like synthetic

Answer 14

=(r^2-1)^3

Answer 15

:D thanks alot. i get it

Answer 16

thats alot of work for one problem lol

Answer 17

haha yea that sure is. im highly obliged

Answer 18

so none of the algebra I did needs explaining?

Answer 19

nope its quite clear :) i completely get it, thanks to you :)

Answer 20

:)

Answer 21

@myininaya. could you please repost your cal0009.pdf attachment. it won't open for me. thanks