Solving Cubic Equations - General Method

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To solve the cubic equation ax^3 + bx^2 + cx + d = 0, the general method involves reducing it to a simpler form without the x^2 term. This is achieved by substituting x with y + y0 to eliminate the quadratic term. The resulting reduced cubic equation can be solved using the relationship x^3 + mx - n = 0, where m and n are derived from the original coefficients. The solutions for a and b can yield up to three distinct values for x, as each cube root provides multiple complex solutions. This method effectively allows for solving all cubic equations systematically.
TheDestroyer
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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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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.
 
Thanks Guyz,

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

I'm very thankful :):)
 
?

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


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).
 
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