## anonymous 5 years ago r^6 - 3r^4 + 3r^2 - 1 = 0 .. How to solve this differential equation ???

1. anonymous

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 $r = \pm \sqrt{x}$

2. anonymous

ohh yea i get it! thanks alot !

3. myininaya

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

4. anonymous

:) 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.

5. myininaya

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

6. myininaya

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

7. anonymous

@myininaya. thanks it sure did help

8. anonymous

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

9. anonymous

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

10. anonymous

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 ?

11. anonymous

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

12. anonymous

(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)

13. myininaya

look i have something else if you didnt like synthetic

14. myininaya

=(r^2-1)^3

15. anonymous

:D thanks alot. i get it

16. myininaya

thats alot of work for one problem lol

17. anonymous

haha yea that sure is. im highly obliged

18. myininaya

so none of the algebra I did needs explaining?

19. anonymous

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

20. myininaya

:)

21. anonymous

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