Solving Cubic Equations - General Method

  • Thread starter TheDestroyer
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In summary, the conversation is about solving cubic equations. The post referenced provides a summary of the general method, which involves using the equation (a-b)^3 + 3ab(a-b) = a^3 - b^3. By substituting values for a and b, the equation can be solved for the variable x. This method can also be applied to general cubic equations.
  • #1
TheDestroyer
402
1
Hi guyz,

How can I solve the equation with the form :

ax^3 + bx^2 + cx + d = 0

I want the general way to solve allllllllll cubic equations..

Thanks
 
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  • #3
The post above is very good but here it is in a nutshell:

We know that (a- b)3= a3-3a2b+ 3ab2- b3 and that 3ab(a-b)= 3a2b- 3ab so:

(a-b)3+ 3ab(a-b)= a3- b3
(the middle terms cancel).

That is: if we pick any two numbers a, b and let x= a- b, m= 3ab and n= a3- b3, then x3+ mx= n.

What about the other way? If we are given m and n, can we solve for a and b (and so find x)?

Yes, we can. From m= 3ab, we have b= m/(3a). Put that into
a3- b3= n and we have a3- m3/(33a3)= n

Multiply both sides of the equation by a3 and we have the (6th degree!) equation
a6- (m/3)3= naa3.

But if we let u= a3, this reduces to a quadratic equation for u: u2- nu- (m/3)3= 0.

We can solve that by the quadratic formula:
u= a3= (n +/- √(n2+ 4(m/3)3))/2= (n/2)+/- √((n/2)2+ (m/3)3).

Since a3- b3= n,
b3= a3- n
= (-n/2)+/- √((n/2)2+ (m/3)3).

Solving for a and b,

a= (((n/2)+/- √((n/2)2+ (m/3)3))1/3
b= ((-(n/2)+/- √((n/2)2+ (m/3)3))1/3

and, finally, x= a-b.


Notice that the equation was x3+ mx- n= 0 which has no x2. This is the "reduced" cubic.

To solve a general cubic, x3+ ax2+ bx+ c= 0,
"Shift" the variable: replace x by y+ y0 in this equation and then choose y0 to make the coefficient of y2 equal to 0.
 
  • #4
Thanks Guyz,

Thanks Guyz, I hope these equations are true, because i can't check them right now :)

I'm very thankful :):)
 
  • #5
?

Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x? or 1 only?
 
  • #6


Originally posted by TheDestroyer
Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x? or 1 only?

Yes. Note that the "a" and "b" in that reply are derived from cube roots. So "a" and "b" each have 3 possible (complex) values giving a total of 9 combinations in the solution (though at most 3 will be distinct).
 
Last edited:

1. What is a cubic equation?

A cubic equation is a polynomial equation of the form ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants and x is the variable.

2. What is the general method for solving cubic equations?

The general method for solving cubic equations is known as the cubic formula, which involves substituting the values of a, b, c, and d into a specific equation and then using algebraic manipulation to find the roots of the equation.

3. Is there a simpler method for solving cubic equations?

Yes, there is a simpler method called the rational root theorem, which states that if a polynomial has rational roots, those roots will be in the form of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

4. Can all cubic equations be solved using the general method?

Yes, all cubic equations can be solved using the general method. However, in some cases, the solutions may involve complex numbers.

5. Are there any real-world applications of solving cubic equations?

Yes, cubic equations are commonly used in physics, engineering, and economics to model various phenomena and solve problems related to these fields.

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