Here, I retrieves my talks in Mathematics community of Orkut.

Shortcut to find complex rootsEdit

2008y 1m 11dEdit


As @Anil posted, your two equations becomes 3rd order and 4th order polynomial solving problems. Hence, you can use root formula for 3rd and 4th polynomial equations. As 2nd order formula has simple root formulas, both 3rd and 5th polynomials also have the root equations, which little bit more complex but not serious. Hence, if you use the root formulas, you will solve your equation with very short time.


2008y 12m 30dEdit


You provided really good greeting statements for new year. If you would have not used red letters too much which makes members hard to read letters, it would be really impressive to me.

The global market crisis is now really problem for all of us. We know that in general a crisis gives opportunity as well but right now, it is really hard to find the hole to see sun shining. Everything are really in dark and looks the darkness never to be cleared up. As a bight day will come after a dark nights is gone. Particularly, whenever in a winter, I really hate it and ask to disappear quick and to come new spring again. Although a spring comes late than we anticipate but anyway it comes to us after a certain amount of period. I believe that the global market crisis will be solved in any form, rather it goes to the unsolvable situation.

Let's wait a moment till the shun shine is bright again!

How can I solve x^3+ax^2+bx+c=0?Edit

2008y 12m 31dEdit

@★яιαz нυѕѕαιη

Thank you for providing a reference describing a solution of the cubic polynomial equation. In my high school period, I tried and tried to solve it by myself since I can solve the square polynomial easily. However, I could not find the solution and not even a suboptimal solution. Could you let me know how Nicol and Scipione can find the solution? Moreover, the history of Mathematics is far earlier than 16c, why no mathematicians could find the solution before?

The "Cubic Formula"Edit

The following describes the solution of the cubic formula. Although I have tried continuously to find the solution when I study in schools, I could not find the solution. It is just one order added to the square formula but the solution is really hard to attack as we know many famous mathematicians failed to find the solution. Because of its complexity, it may be solved later than when it was first solved or even not yet solved. I can know it from a book in a library closed to my home and also received a solution in Orkut community friend ★яιαz нυѕѕαιη נυηєנσ★™. From the source below we show only solution rather than including examples. To see the examples, you may refer to the source below.


Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1510-1576) in his Algebra book Ars Magna.

Our objective is to find a real root of the cubic equation.

$ a x^3 + b x^3 + cx + d = 0 $

The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. The solution proceeds in two steps. First, the cubic equation is depressed; then one solves the depressed cubic.

Depressing the cubic equationEdit

The trick, which transforms the general cubic equation into a new cubic equation with missing $ x^2 $-term is due to Nicolo Fontana Tartaglia (1500-1557). We apply the substitution

$ x = y - \frac{b}{3a} $

to the cubic equation, to obtain:

$ a \left( y - \frac{b}{3a} \right)^3 + b \left( y - \frac{b}{3a} \right)^2 + c \left( y - \frac{b}{3a} \right) + d $

Multiplying out and simplifying, we obtain the depressed cubic

$ a y^3 + \left( c - \frac{b^2}{3a} \right) y + \left( d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0. $

Solving the depressed cubicEdit

We are left with solving a depressed cubic equation of the form

$ y^3 + Ay = B. $

How to do this had been discovered earlier by Scipione dal Ferro (1465-1526). We will find $ s $ and $ t $ so that

$ 3st = A $
$ s^3 - t^3 = B. $

It turns out that y = s - t will be a solution of the depressed cubic. Let's check that: Replacing A, B and y as indicated transforms our equation into

$ (s-t)^3+3st(s-t)=s^3 - t^3. $

This is true since we can simplify the left side by using the binomial formula to:

$ (s^3-2s^2t+3st^2-t^3)+(3s^2t-3st^2)=s^3-t^3. $

How can we find $ s $ and $ t $ satisfying (1) and (2)? Solving the first equation for $ s $ and substituting into (2) yields:

$ \left( \frac{A}{3t} \right)^3 - t^3 = B. $

Simplifying, this turns into the tri-quadratic equation

$ t^6 + Bt^3 - \frac{A^3}{27} $

which using the substituting $ u = t^3 $ becomes the quadratic equation

$ u^2 + Bu^ - \frac{A^3}{27} $

From this we can find a value for u by the quadratic formula, then obtain $ t $, afterwards $ s $ and we're done.

2008y 12m 30dEdit


I'm wonder how I can solve the 3rd order polynomial equation. Is it possible to use derivation or integral to solve it quickly?

Binomial TheoremEdit

2008y 12m 27dEdit


I don't understand the following question. Since (1+x)^n = ... is well-known binomial expansion, we don't need to let it as a assumption. It is just a factor. Moreover the equation of (-1)^n n.2nCn looks unclear. Does it represent '(-1)^n n 2nCn' where 2n = the product of 2 and n? It is so complicated, although it is caused by the limitation of Orkut editor which does not provide the way to display mathematical formulations appropriately.

If (1+x)^n=nC0 + nC1 x + nC2 x^2+...+nCn x^n,then prove that(Co)^2 - 2(C2)^2 + 3.(C3)^2 - ... - 2n.(C2n)^2 = (-1)^n n.2nCn.

See AlsoEdit