Solve equation by completing the cube: x³ - 6x² + 12x - 9 = 0?

Question

I meant real solutions.

I think that one needs to explicitly ask for complex solutions, or at least to operate in terms of 'polynomials' and 'roots' to imply that underlying filed is that of complex numbers.

'Solve eqution' in my oppinion does not imply complex arithmetics.

Answer 1

lol

-6 / 3 = -2 (3 is C(3,1) the binomial coefficient)
12 / 3 = 4 = (-2)^2, so we can, in fact, "complete the cube."

Then (-2)^3 = -8, so that

x^2 - 6x^2 + 12x - 8 = (x-2)^3

Then the equation above becomes

(x-2)^3 - 1 = 0
(x-2)^3 = 1
x-2 = 1
x = 3.

Edit: I assumed that it was a real equation, and thus omitted the complex roots.

Answer 2

Well, the answers so far are almost correct.

(x-2)^3 = x^3 - 6x^2 + 12x - 8
so we have
(x-2)^3 = 1

However, there are three cube roots of every number, not one. The three cube roots of 1 are
1, -1/2 + (√3/2)i, & -1/2 - (√3/2)i

So we have:
x-2 = 1
x = 3
OR
x-2 = -1/2 + (√3/2)i
x = 3/2 + (√3/2)i
OR
x-2 = -1/2 - (√3/2)i
x = 3/2 - (√3/2)i

Answer 3

K-Dub gave the correct soln I suppose. I was reading it up and chanced upon this:

http://mathforum.org/dr.math/faq/faq.cubic.equations2.html

It explains that for x^3 + ax^2 + bx + c = 0, if a^2 - 3b = 0, then the first three terms are the first three terms of a perfect cube, (x+a/3)^3. Fortunately for your case, I didn't have to read the article further :)