Probably an easy question (1 Viewer)

[Giant Lobster]

Is there a way to solve x^3 - 3x^2 + 3x + 3 = 0? (Answer is 1 - 2^(2/3) i think)

Is there some kind of cubic formula (like the quadratic formula we learnt in yr 9) I could use in these situations?
(No im not an idiot, there is one but its not in the syllabus. I remember seeing it before, got a square root within another or somethin and has complex numbers in it I think)

If not, is there a way to do it using agebraic skills? like, rearranging etc...

(Originally this was a question asking me to find the stationary point of y = (x^3 - 3x)/((x-1)^2) but i dunno how to solve cubics without rational roots; perhaps there was another way)

 

[wogboy]

x^3 - 3x^2 + 3x + 3 = 0

if you look closely it resembles the bionomial expansion of (x-1)^3.

(x-1)^3 = x^3 - 3*x^2 + 3x - 1

x^3 - 3x^2 + 3x + 3 = (x-1)^3 + 4
(x-1)^3 = -4
x = 1 - 4^(1/3) = 1 - 2^(2/3)

Highly non trivial question

There is Cardan's formula for solving any cubic polynomial, but it's an awful eyesore to look at :

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

 

[Giant Lobster]

amazing.... gawd Im crap at this

holy crap that cubic formula owns!
*looks for the quartic formula*
...
wow.

edit::
hey thats skilled, how u "completed the cube". Can that method be applied in all cases?
Correct me if im wrong but a little fiddling around after reading ur reply, I dont think it can be applied everywhere like completing the square can. The aim of this "completing the cube" is to get rid of everything but a constant right? hmmm doesnt seem to work with all cubics.

 

[Constip8edSkunk]

in the case above, u know that the cubics will expand out to the likes of 1:3:3:1 for the coefficients. as the coefficient of x^2 is negative you know it would b (x-a)^3 +something so the constant term would b '-1'. this only works cuz the other 3 coefficients fits to the 1:3:3:1 pattern othe wise you'd have something like (ax+b)^3+cx+d

 

[Grey Council]

the only problem is see'ing the solution the first time.

hehe. *dont worry guardian, you'll get better with practise*. Hope so.

And that cubic formula? :O if you memorise it, your crazy. Its 10 times quicker to just try and find the roots some other way. ANY other way.

 

[Giant Lobster]

heh... id say cubics with rational roots can be done by practically anyone. the formula comes in handy when theres all irrational roots / complex roots.

 

[:: ck ::]

holy cow @ the formula

i dont think we'd be allowed to use that in the hsc wud we? is it out of the syllabus

 

[ND]

You can always solve a cubic eqn by using substitutions. In this case the substitution would be x=y+1:

(y+1)^3-3(y+1)^2+3(y+1)+1=0
expanding that out gives:
y^3+4=0

Basically the aim of the subtitution is to get rid of the y^2 (and in this case it also got rid of the y).

 

[Giant Lobster]

yeah this substitution method ive seen in the leet sections of harder books, but they dont really teach you anywhere. I really have to learn how to solve cubics using that method.