All of us have solved many problems on cubic polynomials who have their roots in Arithmetic Progression and Geometric Progression.So I have found a relationship between the coefficients of these polynomials which makes these roots special.

So first , let's start with the roots in Arithmetic Progression.

Consider a polynomial : $\alpha_1x^3+\alpha_2x^2+\alpha_3x+\alpha_4$.

Let its roots be $\beta , \gamma , \lambda$ such that they form an arithmetic progression.So we have a relation :

$\Rightarrow 2\gamma = \beta + \lambda \dots (1)$

Now our favorite polynomial tool cum friend that is , Vieta's formula will help us to have some more relations:

$\sigma_1=\beta +\gamma + \lambda = \dfrac{-\alpha_2}{\alpha_1} \dots (2) \\ \sigma_2=\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \dots (3) \\ \sigma_3= \beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \dots (4)$

Using $(1),(2)$ we have :

$\beta +\gamma + \lambda = 2\gamma+ \gamma = 3\gamma= \dfrac{-\alpha_2}{\alpha_1} \Rightarrow \gamma = \dfrac{-\alpha_2}{3\alpha_1} \dots (5)$

Using $(1),(3),(5)$ we have :

$\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \dots(6) \\ \Rightarrow 2\gamma^2 + \beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow 2\left( \dfrac{-\alpha_2}{3\alpha_1}\right)^2 + \beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \beta\lambda = \dfrac{\alpha_3}{\alpha_1} - \dfrac{2\alpha_2^2}{9\alpha_1^2} \\ \Rightarrow \beta\lambda=\dfrac{9\alpha_1\alpha_3-2\alpha_2^2}{9\alpha_1^2} \dots (7)$

Using $(4) , (7)$ we have :

$\dfrac{\sigma_3}{\beta\lambda} = \gamma \\ \Rightarrow \dfrac{\dfrac{-\alpha_4}{\alpha_1}}{\dfrac{9\alpha_1\alpha_3-2\alpha_2^2}{9\alpha_1^2}} = \dfrac{-\alpha_2}{3\alpha_1} \\ \Rightarrow \left(\dfrac{-\alpha_4}{\alpha_1}\right) \times \left(\dfrac{9\alpha_1^2}{9\alpha_1\alpha_3-2\alpha_2^2}\right)=\dfrac{-\alpha_2}{3\alpha_1} \\ \Rightarrow \dfrac{-9\alpha_1^2\alpha_4}{9\alpha_1\alpha_3-2\alpha_2^2} \\ \Rightarrow 27\alpha_1^2\alpha_4=9\alpha_1\alpha_2\alpha_3 - 2\alpha_2^3$

Thus , we can write the relation between coefficients as :

$\Large\boxed{\alpha_4(2\alpha_2^3+27\alpha_1^2\alpha_4) = 9\displaystyle\prod_{i=1}^4 \alpha_i}$

Now let the roots $\beta , \gamma , \lambda$ form a Geometric progression.Thus we have the relation :

$\gamma^2=\beta\lambda \dots (8)$

Using $(1)$ we have :

$\beta+\lambda = \dfrac{-\alpha_2}{\alpha_1} - \gamma \Rightarrow \beta+\lambda=-\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) \dots (9)$

Using $(4),(8)$ we have :

$\beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \Rightarrow \gamma^3= \dfrac{-\alpha_4}{\alpha_1} \Rightarrow \gamma= \sqrt[3]{ \dfrac{-\alpha_4}{\alpha_1}} \dots (10)$

Using $(6),(9),(10)$ we will have :

$\gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow -\gamma\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) + \gamma^2 = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \dfrac{-\alpha_2\gamma}{\alpha_1} - \gamma^2+\gamma^2 = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \dfrac{-\alpha_2\gamma}{\alpha_1} = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma = \dfrac{-\alpha_3}{\alpha_2} \\ \Rightarrow \sqrt[3]{ \dfrac{-\alpha_4}{\alpha_1}} = \dfrac{-\alpha_3}{\alpha_2} \\ \Rightarrow \dfrac{-\alpha_4}{\alpha_1} = \dfrac{-\alpha_3^3}{\alpha_2^3} \\ \alpha_4\alpha_2^3 = \alpha_1\alpha_3^3$

This when rearranged properly we get the following relationship between the coefficients:

$\Large\boxed{\dfrac{\alpha_1}{\alpha_4}=\left(\dfrac{\alpha_2}{\alpha_3}\right)^3}$

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## Comments

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TopNewestI assume you're only solving for cubic polynomials.

Now try for Harmonic progression as well! And post some challenges!

Nice read! Great work!

Next stop: Apply Newton's sum! ｡◕‿◕｡

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Hey, I observed that even $\left\lceil { \pi }^{ \pi }\ln { \pi } \right\rceil$ is the meaning of life.

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Thanks a lot. I will for sure try it for Harmonic progression. And yeah , I will not make Newton's sums feel lonely. $\ddot\smile$.

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This is really cool. Good work, keep it up! :)

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Thanks a lot :)

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@Calvin Lin I will be happy if you read this. :)

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While the (tedious) algebra work has (hopefully) no mistakes, it ends up obscuring the "true beauty" of the equation. Having found the nice relation, you should think about alternative / better ways of proving the result. Often, the first solution (esp a brute force one) is not the best approach. There will be parts that can be discarded, or rewritten.

For example, let's take the Geometric progression case. Here is a nice simple solution:

Now, I have still hidden the beauty of the solution. If you truly understand it, the following problem has a "one line" solution.

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My thought:Its a good idea to assume the roots as $1,r,r^2$.But one can argue that we get the relationship $\dfrac{\alpha_4}{\alpha_1} = \left(\dfrac{\alpha_3}{\alpha_2}\right)^3$ only if one of the roots is $1$.What about when it is not equal to $1$? There is no guarantee that the relationship remains the same for all possible roots.For a polynomial whose $n$ roots are in geometric progression, I think $\dfrac{\alpha_{n+1}}{\alpha_1} = \left(\dfrac{\alpha_n}{\alpha_2}\right)^n$. Am I right?

Also I didn't understand what do you want me to find about $\dfrac{\alpha_{n-i}}{\alpha_{2+i}}$.

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$a^3$" means. What this is, is that if the polynomial $h(x)$ has roots $a , ar, ar^2$, then the polynomial $h(ax)$ will have roots of $1, r, r^2$. Thus, we can apply the above to conclude about the coefficients of $h(ax)$, from which that gives us information about $h(x)$. It is not immediately apparent that they have to satisfy the same equation, so we have to verify why $h(x)$ still works.

Right, as I said, some steps/explanation is still missing. One of the steps is explaining what "Dividing throughout byOf course, another approach would be to check that

$(x-a)(x-ar)(x-ar^2) = x^3 - (a +ar+ar^2) x^2 + (a^2r + a^2r^2 + a^2r^3) - ar^3.$

from which the identity follows. However, this is still "tedious magical algebra", instead of getting to the "actual reason why this relationship should exist" which would allow us to easily generalize to $a_{n-i} / a_{2+i}$.

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Gud... Thanks...

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Welcome buddy!

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Excellent work!

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gud work nihar

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@Nihar Mahajan @Pi Han Goh What about this?

Let the roots be $x-y,x,x+y$ , by Vieta's relations we have

$3x=\dfrac{-a_2}{a_1}$

But since $x$ is a root of the equation , therefore substitute this value of $x$ in the equation and simplify by to get the desired result.

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No, you're abusing the math notations. You can't have "x" is a root the equation (with variable "x" again). That makes no sense.

You're on the right track, yes, if you let "p-q, p, p + q" be the 3 roots that follow an AP, then yes, you can prove all the results that Nihar has shown with less work.

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Oops...I should have used the same variable..Sorry!!

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Here are some relevant articles: Cardano's method and Cubic discriminant.

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Thanks a lot sir.!

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And even the GP case can be done easily with a similar analysis.

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