Cubic roots

[ali PMPAINT]

Homework Statement

the answer to y^3+py+q=0

Relevant Equations

Picture

So, you can prove it by using (v+c)^3-3vc(v+c)-(v^3+c^3)=0, where v+c=y and -3vc=p and v^3+c^3=q. The problem is, when I try to solve a one, I just get one answer(it should be 3) and I don't know what the article meant by wi. What am I missing?

 

[Mark44]

, post: 6177224, member: 661503"]
Problem Statement: the answer to y^3+py+q=0
Relevant Equations: Picture

View attachment 243310
So, you can prove it by using (v+c)^3-3vc(v+c)-(v^3+c^3)=0, where v+c=y and -3vc=p and v^3+c^3=q.

I think you have a sign error. $-(v^3 + c^3)$ should be $+(v^3 + c^3)$ according to the substitution you show

The problem is, when I try to solve a one, I just get one answer(it should be 3) and I don't know what the article meant by wi. What am I missing?

It says that $w_i$ is one of the cube roots of unity (1). In polar form these are
$w_0= 1$
$w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$
$w_2 = \cos(\frac {4\pi}3) + i\sin(\frac {4\pi}3)$
Notice that two of these roots are complex.
If you raise any of these to the 3rd power, you get 1.

 

[SammyS]

So, you can prove it by using (v+c)^3-3vc(v+c)-(v^3+c^3)=0, where v+c=y and -3vc=p and v^3+c^3=q.

I think you have a sign error. −(v3+c3)−(v3+c3)-(v^3 + c^3) should be +(v3+c3)+(v3+c3)+(v^3 + c^3) according to the substitution you show.

... or that should say: $\ -(v^3+c^3) = q \, .$

Also, for the casual reader, this expression, given by OP, is an identity : $\ (v+c)^3-3vc(v+c)-(v^3+c^3)=0\, .$

 

[ali PMPAINT]

I think you have a sign error. $-(v^3 + c^3)$ should be $+(v^3 + c^3)$ according to the substitution you show

Yes, thanks for correcting me.

It says that $w_i$ is one of the cube roots of unity (1). In polar form these are
$w_0= 1$
$w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$
$w_2 = \cos(\frac {4\pi}3) + i\sin(\frac {4\pi}3)$
Notice that two of these roots are complex.
If you raise any of these to the 3rd power, you get 1.

So, I am now familiar a lot with polar form, so if we got an answer like 1 for example, do you set it equal to e^(ix)=cos(x)+isin(x) ?

 

[Mark44]

Yes, thanks for correcting me.


So, I am now familiar a lot with polar form, so if we got an answer like 1 for example, do you set it equal to e^(ix)=cos(x)+isin(x) ?

Here x is 0, so $e^{i0} = \cos(0) + i\sin(0) = 1 + i0 = 1$

 

[ali PMPAINT]

Here x is 0, so $e^{i0} = \cos(0) + i\sin(0) = 1 + i0 = 1$

Yes, but what about the other two answers?
For example, how to get all solutions for x^3+2x-3=0 ?

 

[StoneTemplePython]

Yes, but what about the other two answers?
For example, how to get all solutions for x^3+2x-3=0 ?

do you know what synthetic division is?

 

[Mark44]

Yes, but what about the other two answers?
For example, how to get all solutions for x^3+2x-3=0 ?

You included the formula in your first post of this thread. I gave you all three roots of unity in my previous post. Just use that formula, once for each of $w_0, w_1, w_2$.

 

[SammyS]

It says that $w_i$ is one of the cube roots of unity (1). In polar form these are
$w_0= 1$
$w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$
$w_2 = \cos(\frac {4\pi}3) + i\sin(\frac {4\pi}3)$
Notice that two of these roots are complex.
If you raise any of these to the 3rd power, you get 1.

So, I am now familiar a lot with polar forms, so if we got an answer like 1 for example, do you set it equal to e^(ix)=cos(x)+isin(x) ?

Here x is 0, so $e^{i0} = \cos(0) + i\sin(0) = 1 + i0 = 1$

Yes, but what about the other two answers?

You can get $w_1$ by letting x = 2π/3 in the polar form you gave, which was: e^(ix)=cos(x)+isin(x)

Then $w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$ which is the same as $\displaystyle e^{2\pi i / 3}$

It then follows that $\displaystyle {w_1}^3=\left( e^{2\pi i / 3} \right) ^3 = e^{2\pi i } =1$

 

[ali PMPAINT]

You can get $w_1$ by letting x = 2π/3 in the polar form you gave, which was: e^(ix)=cos(x)+isin(x)

Then $w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$ which is the same as $\displaystyle e^{2\pi i / 3}$

It then follows that $\displaystyle {w_1}^3=\left( e^{2\pi i / 3} \right) ^3 = e^{2\pi i } =1$

Thanks, but now I'm facing another problem. When I tried to solve for x^3+2x-3=0 I got:

I calculated it with a calculator and I got 1 out of this mess(which is true). No matter what I do, I can't simplify the expresion to get 1. And, I found that the other two roots are number 2 (by both calculator and another methods), but I tried to multiply it with $w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$ with one, but I got a wrong result (-1/2+i 3^(0.5)/2 ) which is wrong. Where am I making a mistake?
I am sorry if I am asking a lot, but I really want to know how to solve cubic equations by radicals.

 

[Mark44]

Thanks, but now I'm facing another problem. When I tried to solve for x^3+2x-3=0 I got:
View attachment 243371
I calculated it with a calculator and I got 1 out of this mess(which is true). No matter what I do, I can't simplify the expresion to get 1. And, I found that the other two roots are number 2 (by both calculator and another methods),

Yes, 1 is a solution of the equation $x^3 + 2x - 3 = 0$, because if you replace x by 1, you get 0 = 0, which is a true statement.
However, 2 is not a solution of this equation, despite what your calculator and another method imply. The reason is that if you replace x in the equation by 2, you get $2^3 + 2(2) - 3 = 9 \ne 0$.

Knowing that x = 1 is a solution, that means that (x - 1) is a factor of $x^3 + 2x - 3$. If you divide the cubic polynomial by (x - 1), either by long division or by synthetic division that @StoneTemplePython mentioned, you get a quadratic polynomial that you can break down using the Quadratic Formula. The other two solutions of your equation are complex.

but I tried to multiply it with $w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$ with one, but I got a wrong result (-1/2+i 3^(0.5)/2 ) which is wrong. Where am I making a mistake?
I am sorry if I am asking a lot, but I really want to know how to solve cubic equations by radicals.

 

[SammyS]

Thanks, but now I'm facing another problem. When I tried to solve for x^3+2x-3=0 I got:
View attachment 243371
I calculated it with a calculator and I got 1 out of this mess(which is true). No matter what I do, I can't simplify the expresion to get 1. And, I found that the other two roots are number 2 (by both calculator and another methods), but I tried to multiply it with $w_1 = \cos(\frac {2\pi}3) + i\sin(\frac {2\pi}3)$ with one, but I got a wrong result (-1/2+i 3^(0.5)/2 ) which is wrong. Where am I making a mistake?
I am sorry if I am asking a lot, but I really want to know how to solve cubic equations by radicals.

Yes. the above expression gives you 1 (unity).

It looks like the given expression is incorrect.
The correct results for the two complex solutions to $x^3+2x-3=0$ are $\displaystyle x= \frac{-1 \pm i \sqrt{11}\,}{2} \,.$

A correct form for the solution to the cubic equation: $x^3+px+q=0$ is:

$\displaystyle w_k \left({\sqrt[{3}]{-{q \over 2}+{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}+ w_k{\sqrt[{3}]{-{q \over 2}-{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}} }\ \right) \ , \\ \displaystyle \text{where } w_k=e^{2k\pi i /3} \ .$

Note that for $k \in \{1,\,2,\,3\,\},\ \ w_k$ is one of the three cube roots of unity.

The above may be expressed in a variety of forms.

 

[ali PMPAINT]

Oh yes, my bad, apparently I calculated the roots of x^3-3x+2=0 instead of x^3+2x-3=0 and missed the sign (don't blame me, they are too similar)

A correct form for the solution to the cubic equation: $x^3+px+q=0$ is:

$\displaystyle w_k \left({\sqrt[{3}]{-{q \over 2}+{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}+ w_k{\sqrt[{3}]{-{q \over 2}-{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}} }\ \right) \ , \\ \displaystyle \text{where } w_k=e^{2k\pi i /3} \ .$

So, I think that I had a mistake in my proof, and it doesn't give me a correct answer for more than two answers:

Where is my mistake?