# How to solve this?

TheDestroyer
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

imhereyeah
Homework Helper
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:

Since a3- b3= n,
b3= a3- n

Solving for a and b,

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.

TheDestroyer
Thanks Guyz,

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

I'm very thankful :):)

TheDestroyer
?

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