## shubhamsrg 3 years ago TUTORIAL : on how to solve a 3 degree eqn in x using trigonometry .. i learnt this method quite some time ago and maybe everyone should give this a reading..

1. shubhamsrg

we have : ax^3 + bx^2 + cx+ d =0 for simplification, i like to keep coefficient of x^3 = 1 for that,,lets divide both sides by a (given a not equal to 0) lets say b/a =p , c/a = q , d/a =r => x^3 + px^2 + qx +r =0 what we aim here firstly is to convert this into depressed form ( make co-efficient of 2nd highest degree =0) lets substitute x = (y + m) , for some new variable y and a constant m , we will understand the purpose of doing this soon.. => (y+m)^3 + p(y+m)^2 + q(y+m) + r =0 =>y^3 + y^2(p + 3m) + y(2m + 3m^2 +q) + (m^3 + pm^2 + qm +r) =0 as i said earlier,we want p+ 3m =0 => m= -p/3 thus on making that substitution, we'll get y^3 + Ay + B = 0 where A and B are some known constants.. now if we calculate y , we can easily calculate x as x= y+m or x= y -p/3 here comes the trignometric part: this glorious formulla will do wonders for us as you'll find below : sin 3@ = 3sin @ - 4sin^3 @ (@ denotes theta) so we have y^3 + Ay + B=0 let y = t sin@ for some constant t , => t^3 sin^3 @ + At sin@ + B =0 lets multiply both sides by -4/t^3 => -4 sin^3 @ + -4A sin@ /t^2 + -4B/t^3 = 0 we have to chose t such that coefficient of sin@ = 3 => -4A/t^2 = 3 =>t = sqrt( -4A/3) and -sqrt( -4A/3) on making that substitution, we see we have -4sin^3 @ + 3sin@ =C for some new constant C. => sin3@ = C =>3@ = sin^-1 (C) @ = 1/3 * sin^-1 (C) since you know the value of @, you then know value of y = t sin@ and then x = y+m

2. shubhamsrg

all this might seem tedious and long at first,,but this is very advantageous to know.. this one yields both complex roots as well as real roots..2 of them .. (since you get 2 values of t) hope that helps..

3. shubhamsrg

another thing which we can make out from here that, in any eqn like x^n + px^(n-1) + qx^(n-2) ......... + constant term =0 to make it depressed eqn , we can always substitute x = y - (p/n)

4. hba

What is this method called ?

5. shubhamsrg

i dont know/remember.. :|

6. hba

k

7. hba

3x^3+2x^2+x+1=0 Show me how to solve this step by step please.

8. shubhamsrg

you do it and i shall add my words in between when required..

9. hba

let me start then :)

10. hba

Do i have to suppose x^3=1 ?

11. shubhamsrg

nops..i meant coefficient of x^3 should be made 1,,just for simplification.. so you may divide both sides by 3

12. hba

Dividing the whole equation by 3 x^3+2/3 x^2+1/3x+1/3=0

13. hba

What next ?

14. shubhamsrg

x = y - p/3 .. make that substitution,, p = 2/3 here..

15. hba

p=2/3,q=1/3,r=1/3 I did not get the substitution part lol ? Do you think this method is more effective then the others ?

16. shubhamsrg

i cant say that,,since i dont know other methods! :P nevermind,,whats the problem is substituting ?

17. shubhamsrg

like in x+2 = y, you can substitute directly x=2 ,, similarly,,substitute x= y - p/3

18. shubhamsrg

i'll give a kick.. x^3+2/3 x^2+1/3x+1/3=0 (y - 2/9)^3+ 2/3 (y- 2/9)^3 + 1/3 ( y- 2/9) + 1/3 = 0 did you follow?

19. hba

Ok so we have just done a simple substitution here x=y-2/9.

20. hba

Do i expand them using formula ?

21. shubhamsrg

yep..expand and simplify .. write what you get .. sorry am too lazy to check your work or do it myself..so be careful..please dont mind. :P

22. hba

Anyways i understood the part where i was confused,Thanks a lot.

23. shubhamsrg

so,,you're comfortable with the rest ?

24. hba

Yeah i was confused in this part lol :D

25. shubhamsrg

kk.glad i could help..

26. Hero

Is there a textbook or video tutorial I can find this in?

27. hba

If i remember the name of the method i could have told you :(

28. shubhamsrg

well one of my seniors had told me this..i really cant tell you any name or any vids or theories about it..sorry.. anyways,,are you stuck somewhere ?

29. hba

@Hero This method does not intrest me :) Its very long.

30. shubhamsrg

sorry buddy,,too lazy doing all that..though i could help you ..but giving examples..nah,,not gonna do that..

31. shubhamsrg

ZZZZZzzz...

32. hba

33. shubhamsrg

i dont regret that..i made an effort..and it was satisfactory for me..i dont mind your foolish reviews since you ofcorse,dont get it..

34. hba

@shubhamsrg If you give an example maybe you would be getting more medals think about that :D

35. shubhamsrg

ohh yes,,the medals..it was always my dream to earn more and more medals on open study.. !! o.O

36. hba

Oh come on just provide us with an example please ?

37. hba

If you feel lazy do it after sometime :)

38. shubhamsrg

nevermind,,not gonna do that..if you dont get it..put up your query here..i may help/improvise// but i hate doing hard work more often than not..

39. RajshikharGupta

it is tooo lengthy but indeed beautiful we use to call it 'MAJDOORI'

40. shubhamsrg

we?? o.O who all do you represent? :P

41. shubhamsrg

alright,,seems its lengthy.. in that case,,gimme some time,,i'll post a general formula to the value of x directly..

42. sirm3d

except for the trigonometric substitution, solving a cubic equation appeared in Ars Magna of Girolamo Cardamo, or Cardan.

43. shubhamsrg

here's the general formula for solution of x in x^3 + px^2 + qx + r =0 using trig. substitution : |dw:1354008782695:dw|

44. shubhamsrg

where: m = sqrt(-4A /3) and -sqrt(-4A/3) A = q- (p^2 /3) B = (2p^3 /27) + (qp /3) + r

45. shubhamsrg

hope that aids to the lengthiness ..

46. shubhamsrg

sorry, correction : that'll be -p/3 not +p/3 there in the drawing.. rest all the same..

47. inkyvoyd

cardano is better. except casus irriduciblis. Then this is.

48. shubhamsrg

ohh..i just googled over!! that was amazing! thanks @inkyvoyd also,,just found out vieta's substitution method..too good..probably the easiest..

49. Mathmuse

Haha, reading about this and found this excerpt: The long-sought solution of the general cubic was found, in 1535 by Niccol`o Tartaglia (c. 1500-57). This achievement brought him great celebrity, and he spent the next 10 years visiting the crownedheads of Europe and solving their cubics for them. However, he was persuaded to divulge his secret, on the promise of complete conﬁdentiality, by Girolamo Cardano (1501–76), who promptly published it in his algebra book The Great Art. There followed an acrimonious dispute between Tartaglia and Cardano which preoccupied much of Tartaglia’s later life. At one point, the two agreed to a public duel, in which they would each bring along their favourite mathematical problems for the other to solve. However, it never took place @Hero http://www.admissionstests.cambridgeassessment.org.uk/adt/digitalAssets/110501_Advanced_Problems_in_Mathematics.pdf question #12 has an example of this.

50. shubhamsrg

that was interesting @mathmuse ..thanks for sharing

51. mukushla

nice one :)

52. shubhamsrg

aha..thanks..your appreciations means a lot to me sir.. :)