Let $f(x)=ax^3+bx^2+cx+d$, be a polynomial function, find relation between $a,b,c,d$ such that it's roots are in an arithmetic/geometric progression. (separate relations)

So for the arithmetic progression I took let $\alpha = x_2$ and $r$ be the ratio of the arithmetic progression.

We have:

$$x_1=\alpha-2r, \quad x_2=\alpha, \quad x_3=\alpha +2r$$


$$x_1+x_2+x_3=-\frac ba=3\alpha$$ $$x_1^2+x_2^2+x_3^2 = 9\alpha^2-2\frac ca \to 4r^2=\frac {b^2-3ac}{3a^2}$$ $$x_1x_2x_3=\alpha(\alpha^2-4r^2)=-\frac da$$

and we get the final result $2b^3+27a^2d-9abc=0$.

How should I take the ratio at the geometric progression for roots?

I tried something like

$$x_1=\frac {\alpha}q, \quad x_2=\alpha, \quad x_3=\alpha q$$

To get $x_1x_2x_3=\alpha^3$ but it doesn't really work out..


I have to choose from this set of answers:

$$\text{(a)} \ a^2b=c^2d \quad\text{(b)}\ a^2b^2=c^2d \quad\text{(c)}\ ab^3=c^3d$$

$$\text{(d)}\ ac^3=b^3d \quad\text{(e)}\ ac=bd \quad\text{(f)}\ a^3c=b^3d$$


Using your notations $$ax^3+bx^2+c x+d=a(x-\frac \alpha q)(x-\alpha)(x-\alpha q)$$ Expand the rhs to get after simplifications $$a x^3 -\frac{a \alpha \left(q^2+q+1\right)}{q}x^2+\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}x-a \alpha ^3$$ Compare the coefficients to get $$b=-\frac{a \alpha \left(q^2+q+1\right)}{q}$$ $$c=\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}$$ $$d=-a \alpha ^3$$


The condition you want is $\;D:=(x_1^2-x_2x_3)(x_2^2-x_1x_3)(x_3^2-x_1x_2)=0.$ This can be written as $\;D=e_1^3e_3-e_2^3\;$ where $\;e_1,e_2,e_3\;$ are the elementary symmetric functions of the roots. This simplifies in terms of the polynomial coefficients to $\;ac^3-b^3d\;$ since $e_1=-\frac{b}{a},e_2=\frac{c}{a},e_3=-\frac{d}{a}.$ Thus the answer choice is (d).

By the way, the Special Algebraic Identity id3_3_2_6a $$\;(x_1^2-x_2x_3)(x_2^2-x_1x_3)(x_3^2-x_1x_2)=(x1+x_2+x_3)^3x_1x_2x_3-(x_1x_2+x_1x_3+x_2x_3)^3\;$$ is given in H. S. Hall and S. R. Knight, Higher Algebra, 1957 (p. 439, Examples, 23).


To sum up, a cubic has its roots in arithmetic progression if and only if the arithmetic mean of the roots is a root of the cubic ( so equals one of the roots).

For the geometric progression, we could use the same trick, and say that the geometric mean of $x_1$, $x_2$, $x_3$ is a root of $P$. Alternatively, to avoid cubic roots, one considers the equivalent statement that $x_1x_2x_3$ must equal one of the $x_i^3$. So we set up the equation $Q=0$ with roots $x_1^3$, $x_2^3$, $x_3^3$ and impose the condition that $x_1x_2x_3$ is a root.

The equation for $x_i^3$ can be obtained readily by eliminating $x$ from the equalities $y = x^3$, $ax^3 + b x^2 + c x + d$. We get a cubic equation for $y$ $$a^3 y^3+ (3 a^2 d - 3 a b c + b^3)y^2 +(3 a d^2 - 3 b c d + c^3)y+ d^3=0$$ and the condition is that $-\frac{d}{a}$ is a root of this equation.

In general, given a polynomial $P$ of degree $n$, one can get the condition on the coefficients so that for some ordering of the roots we have an algebraic condition $F(x_1, \ldots, x_n)=0$. We take all the possible permutations of $F(x_{\sigma(1)}, \ldots, x_{\sigma(n)})$ of $F$. The condition is that the product of all these permutations is $0$.

If we want $m$ conditions $F_1= \cdots = F_m=0$, we set up $F=\sum t_i F_i$, where $t_1$, $\ldots t_m$ are variables, consider all the possible permutations of $F$. The condition is that the product of all these is $0$, as a polynomial in $t_1$, $\ldots t_m$.


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