# Reduced cubic equation

1. Sep 6, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

Find the solutions of:

$$x^3-3x-2=0$$

using the Cardano's method.

2. Relevant equations

3. The attempt at a solution

$$x=u+v$$

$$(u+v)^3-3(u+v)-2=0$$

$$u^3+v^3+(u+v)(3uv-3)-2=0$$

$$3uv-3=0$$

$$uv=1$$

$$u^3v^3=1$$

$$u^3+v^3=2$$

$$u^3=v^3=1$$

Now $$u=v=\sqrt[3]{1}$$.

I found u and v with $$z=\sqrt[3]{1}$$ (using complex numbers).

$$u=v=1$$

$$u=v=\frac{-1}{2}+i\frac{sqrt{3}}{2}$$

$$u=v=\frac{-1}{2}-i\frac{sqrt{3}}{2}$$

x=u+v

$$x_1=2$$

$$x_2=-1+i\sqrt{3}$$

$$x_3=-1-i\sqrt{3}$$

I substitute above and something is wrong. Where is the error?????

2. Sep 6, 2008

### anantchowdhary

there may be a better way out..observinf that if $$f(x)=x^3-3x-2$$.
clearly $$f(-1)=0$$
satisfies the eqn...hence
$$(x+1)$$ is a factor of $$f(x)$$.
Now use division and get $$f(x)$$ in the form of factors and put the other factor(other than $$x+1$$)
to be zero and solve for
$$x$$

3. Sep 6, 2008

### Physicsissuef

Yes, I know that there are few ways, but I need to solve it with Cardanos. Please help.

4. Sep 6, 2008

### HallsofIvy

Staff Emeritus
?? I don't see how this follows directly, although it is true. Since you have defined u and v such that uv= 1, you have v= 1/u and so $u^3+ 1/u^3= 2$. Multiplying both sides by u3, you get the quadratic (in u3) (u3)2- 2(u3)+ 1= 0 which does give u3= 1 or u3= -1 as roots.

No. This is your error. It does not follow from u3= v3 that u= v. Only that they are both cube roots of 3. And, as you note below, there are 3 of those.

You do NOT know, as I said, that u= v. You know, rather, that uv= 1 so v= 1/u.
If
$$u= \frac{-1+ i\sqrt{3}}{2}$$
then
$$v= \frac{2}{-1+ i\sqrt{3}}$$
rationalize the denominator by multiplying both numerator and denominator by $-1+ i\sqrt{3}$ and you get
$\frac{-1- i\sqrt{3}}{2}$
the other non-real root of z3= 1.

Now, x= u+ v gives x= -1 which is correct. The three roots of x3- 3x- 2= 0 are 2, -1, -1.

5. Sep 7, 2008

### Physicsissuef

So, first I found the 3 solutions of u , and then substitute in uv=1, I get v and then substitute in x=u+v, right?

6. Sep 7, 2008

### HallsofIvy

Staff Emeritus
Yes, basically, that is what you do.

7. Sep 7, 2008

### Physicsissuef

Ok, thanks. But why I read that also suming $u_1+v_1=x_1[/tex], will work if I found [itex]v_1$, by using complex algebra?