# Cubic formula?

1. Feb 9, 2008

### Bill Foster

The quadratic formula is, as you are well aware:

$$ax^2+bx+c=0$$ $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

What is the cubic formula?

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

Thanks.

2. Feb 9, 2008

### jhicks

Cardano's formula will get you one root, but you have to transform the general equation a*x^3+b*x^2+cx+d=0 into e*z^3+f*z+g=0 by use of the substitution z=x-b/3. Mathworld explains a bit more: http://mathworld.wolfram.com/CubicFormula.html

3. Feb 9, 2008

4. Feb 9, 2008

### Bill Foster

Thanks.

5. Oct 17, 2010

### kennysmith39

could someone just show me with an example on how to work the cardano and Del ferro method from start to finish say using x^3-6x^2+2x-1

6. Oct 17, 2010

### HallsofIvy

Staff Emeritus
The first thing you do is "reduce" the equation to the form $x^3+ bx= c$ without any $x^2$ term. To do that, let x= y- a. Then $x^3= (y- a)^3= y^3- 3ay^2+ 3a^2y- a^3[/math] and $x^2= (y- a)^2= y^2- 2ay+ a^2[/math] $$x^3- 6x^2+ 2x-1= y^3 -ay^2+ a^2y- a^3- 6y^2+ 12ay- 6a^2+ 2y- 2a- 1$$$$= y^3+ (-a- 6)y^2+ (a^2+ 2)y+ (-a^3- 2a- 1)[/math] That will have no "[itex]y^2$" term is a= -6 and, in that case, the polynomial is $y^3+ 38y+ 229$ and so our equation is $y^3+ 38y+ 229= 0$. Here's a quick review of Cardano's formula: $(a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$ $-3ab(a+ b)= -3a^2b- 3ab^3$ so that $(a+ b)^3+ 3ab(a+ b)= a^3+ b^3$. In particular, if we let x= a+b, m= 3ab, and $n= a^3+ b^3$, $x^3+ mx= n$. Can we go the other way? That is, given m and n, can we find a and b and so find x? Yes, we can. From $m= 3ab$, we have $b= m/(3a)$ so $a^3+ b^3= a^3+ m^3/(3^2a^3)= n$. Multiplying through by $a^3$, $(a^3)^2+ m^3/3^3= na^3$ or $(a^3)^2- na^3+ m^3/3^3= 0$. We can think of that as a quadratic equation in $a^3$ and solve it with the quadratic formula: [tex]a^3= \frac{n\pm\sqrt{n^2- 4\frac{m^3}{3^3}}}{2}$$= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}$$ and a is the cube root of that. From $n= a^3+ b^3$ we have $$b^3= n- a^3= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}$$. Now, in this problem, $y^3+ 38y+ 229= 0$ or $y^3+ 38y= -229$ so m= 38 and n= -229. $$\frac{n}{2}= -\frac{229}{2}$$ and $$\left(\frac{n}{2}\right)^2= \frac{52441}{4}$ [tex]\frac{m}{3}= \frac{38}{3}$$
and
$$\left(\frac{m}{3}\right)^3= \frac{54872}{27}$$

So
$$a^3= -\frac{229}{2}\pm\sqrt{\frac{52441}{4}- \frac{54872}{27}}$$
and
$$b^3= -\frac{229}{2}\mp\sqrt{\frac{52441}{4}- \frac{54872}{27}}$$

Calculate those numbers, take the cube roots to find a and b and then find x= a+ b.
Each of those will have 3 cube roots but in the various ways of combining them, some things will cancel so that there will be, at most, 3 roots to the equation.

7. Oct 17, 2010

### kennysmith39

thank yo for taking the time to show me this but im still confused, after plugging in y-a into the equation how does a= -6