Given that $x_1, x_2, x_3$ are the roots of the polynomial $x^3-2x^2+3x+5=0$ find $(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$.

Question

Consider the polynomial:

$$x^3-2x^2+3x+5=0$$

where $x_1, x_2$ and $x_3$ are the roots of the above polynomial. Now, consider the following determinant, which is defined using the above given roots:

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \\ \end{vmatrix}$$

And what is asked of me is to find $\Delta^2$.

After a bit of manipulation I found the following:

$$\Delta = (x_2-x_1)(x_3-x_1)(x_3-x_2)$$

Interestingly enough, this type of matrix has a special name: Vandermonde matrix and instead of doing that bit of manipulation after which I arrived at the above expression for $\Delta$, I could've used the formula given on that wikipedia page. Anyways...

So, I have to find:

$$\Delta^2 = (x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$$

The problem is that I cannot find any of the roots. I used the rational root theorem and found that there are no rational roots. None of the divisors of the free term, $5$, give $0$ when plugged into the polynomial. I tried all options: $\{\pm 1, \pm 5 \}$ and they all give something $\ne 0$.

So then I used the notation:

$$f(x) = x^3-2x^2+3x+5$$

found the derivative:

$$f'(x) = 3x^2-4x+3$$

and I observed that $f'(x) > 0$ for all $x \in \mathbb{R}$. So the function $f$ is strictly increasing, so we can have at most one solution to $f(x) = 0$. Because of what I showed above, this solution cannot be rational. So I concluded that we have one rational root and two complex (and conjugate, since $f \in \mathbb{R}[X]$) roots. But this is as far as I got. I cannot find them. And I tried finding $\Delta^2$ without finding the roots, but I couldn't solve that either.

Answer 1

The 'easy way' to do this leans on power sums instead of elementary symmetric polynomials. We are implicitly working in $\mathbb C$ for this problem.

For any degree $n$ monic polynomial, first encode it in an $n$ x $n$ Companion matrix $C$.

now consider the power sum for $k\in\{1,2,3,...,\}$
$s_k := \lambda_1^k + \lambda_2^k +....+ \lambda_n^k = \text{trace}\big(C^k\big)$ and
$s_0:= n$
and $\lambda_i$ are the roots to your polynomial / the eigenvalues of $C$

now consider the matrix
$M_n := \begin{bmatrix} s_0 & s_1 & s_2 & \cdots & s_{n-1}\\ s_1 & s_2& s_3 & \cdots & s_n \\ s_2& s_3 & s_4 & \cdots & s_{n+1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ s_{n-1} & s_{n} & s_{n+1} & \cdots & s_{2n-2} \end{bmatrix}$
for your problem here it is just

$M_3 := \begin{bmatrix} s_0 & s_1 & s_2\\ s_1 & s_2& s_3\\ s_2& s_3 & s_4 \\ \end{bmatrix}$
(note: matrix multiplication is only needed to get $s_2$. You get $s_0$ and $s_1$ immediately. Also $s_3$ and $s_4$ are obtainable by Cayley Hamilton.)

and
$\det\big(M_3\big) = \Delta^2 = (\lambda_2-\lambda_1)^2(\lambda_3-\lambda_1)^2(\lambda_3-\lambda_2)^2$

because
$M= V^TV \longrightarrow \det\big(M\big)=\det\big(V^TV\big)=\det\big(V^T\big)\det\big(V\big)=\det\big(V\big)^2$
(note that is a transpose, not conjugate transpose. This factorization turns out to be quite useful.)

where, for avoidance of doubt, $V$ is the Vandermonde matrix, shown below for the $n=3$ case
$V := \begin{bmatrix} 1 & \lambda_1 & \lambda_1^2 \\ 1 & \lambda_2& \lambda_2^2\\ 1 & \lambda_3 & \lambda_3^2 \\ \end{bmatrix}$

Answer 2

$\Delta$ is not symmetric, but $\Delta^2$ is, so it can be expressed in terms of $a=x_1+x_2+x_3$, $b=x_1x_2+x_2x_3+x_3x_1$ and $c=x_1x_2x_3$. Indeed, we have:

$$\Delta^2 = a^2 b^2 + 18 abc - 4 b^3 - 4 a^3 c - 27 c^2$$

The simplest way I know to prove this identity, is like this: let $x=x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $y=x_1x_2^2+x_2x_3^2+x_3x_1^2$. Then:

$$\Delta^2=(x-y)^2=(x+y)^2-4xy$$

It's pretty simple to notice that $x+y=ab-3c$ and for $xy$, expanding:

$$xy=c(x_1^3+x_2^3+x_3^3)+(x_1^3x_2^3+x_2^3x_3^3+x_3^3x_1^3)+3c^2$$

and for the sum of cubes we have the well-known factorization:

$$x_1^3+x_2^3+x_3^3 = 3c+a(a^2-3b)$$

and similarly:

$$x_1^3x_2^3+x_2^3x_3^3+x_3^3x_1^3=3c^2+b(b^2-3ca)$$

Replacing back all of this:

$$ \begin{aligned} xy &= c[3c+a(a^2-3b)]+[3c^2+b(b^2-3ca)]+3c^2\\ &= b^3 - 6 a b c + 9 c^2 + ca^3 \end{aligned} $$

and thus:

$$ \begin{aligned} \Delta^2 &= (ab-3c)^2-4(b^3 - 6 a b c + 9 c^2 + ca^3 )\\ &= a^2b^2+18abc-4b^3-4a^3c-27c^2 \end{aligned} $$

And we can determine $a,b,c$ from Vieta's ($a=2, b=3, c= -5$). In the end $\Delta^2=-1127$.

Answer 3

For a cubic equation $ax^3+bx^2+cx+d=0$,

the roots $x_1, x_2, x_3$ have the following properties:

$x_1+x_2+x_3=-\frac ba$

$x_1x_2+x_1x_3+x_2x_3=\frac ca$

$x_1x_2x_3=-\frac da$

Since you have $x^3-2x^2+3x+5=0$,

$x_1+x_2+x_3=2$

$x_1x_2+x_1x_3+x_2x_3=3$

$x_1x_2x_3=-5$

Can you do the rest?

Answer 4

Let $x_1+x_2+x_3=3u$, $x_1x_2+a_1x_3+x_2x_3=3v^2$ and $x_1x_2x_3=w^3$.

Thus, $$u=\frac{2}{3},$$ $$v^2=1,$$ $$w^3=-5$$ and $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=$$ $$=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=$$ $$=27\left(\frac{4}{3}-4+\frac{160}{27}-20-25\right)=-1127.$$

Answer 5

Since $x_1$, $x_2$, $x_3$ are the roots of the polynomial $f(x)=x^3 + a x^2 + b x + c$, they are the eigenvalues of the companion matrix $$A=\left(\begin{matrix} 0 & 0 & -c \\ 1 & 0 & -b\\ 0 & 1 & -a \end{matrix} \right)$$

Now we have $$f(x) = (x-x_1)(x-x_2)(x-x_3)$$ so $$f'(x) = (x-x_2)(x-x_3) + (x-x_1)(x-x_3) + (x-x_2)(x-x_3)$$ and from this we get \begin{eqnarray} f'(x_1) = (x_1-x_2)(x_1-x_3)\\ f'(x_1) = (x_2-x_1)(x_2-x_3)\\ f'(x_3) = (x_3-x_1)(x_3-x_2) \end{eqnarray} and so $$\Delta = (x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = - f'(x_1) f'(x_2)f'(x_3)$$

Now, $f'(x_i)$, $i=1,3$ are the eigenvalues of the matrix $f'(A)$. We conclude

$$\Delta = - \det f'(A) = - \det ( 3 A^2 + 2 a A + b I_3)= - \left| \begin{matrix} b & -3 c & a c \\ 2a & -2 b & a b - 3 c \\ 3 & - a& a^2 - 2 b \end{matrix} \right |$$ and so the discriminant is $$\Delta = - 4 a^3 c + a^2 b^2 + 18 a b c - 4 b^3 - 27 c^2$$

Answer 6

Given $x_1,x_2,x_3$ as roots of \begin{align}x^3-2x^2+3x+5&=0 \tag{1}\label{1} \\ \text{and }\quad \Delta &= (x_1-x_2)(x_2-x_3)(x_3-x_1) \tag{2}\label{2} \end{align} find $\Delta^2$.

Surprisingly, the Ravi substitution works smoothly in this case, despite that $x_1,x_2,x_3$ are complex numbers rather than positive reals, as well as the "sides" $a,b,c$, "semiperimeter" $\rho=\tfrac12(a+b+c)$, "inradius" $r$ and "circumradius" $R$ of the corresponding "triangle" are also most possibly complex.

So, using a substitution

\begin{align} x_1&=\rho-a \tag{3}\label{3} ,\\ x_2&=\rho-b \tag{4}\label{4} ,\\ x_3&=\rho-c \tag{5}\label{5} ,\\ \text{and }\quad x&=\rho-y \tag{6}\label{6} \end{align}

in \eqref{1}, we have a cubic equation

\begin{align} y^3+(-3\rho+2)\,y^2+(3\rho^2-4\rho+3)\,y-\rho^3+2\rho^2-3\rho-5&=0 \tag{7}\label{7} \end{align}

which roots are $a,b,c$.

Given that $a,b,c$ are also the roots of the cubic

\begin{align} y^3-2\rho\,y^2+(\rho^2+r^2+4\,r\,R)\,y-4\rho\,r\,R&=0 \tag{8}\label{8} , \end{align}

we can trivially find that we must have $\rho=2$, thus \eqref{7}, \eqref{8} are simplified to \begin{align} y^3-4y^2+7y-11&=0 \tag{9}\label{9} ,\\ y^3-4y^2+(4+r^2+4rR)\,y-8\,r\,R &= 0 \tag{10}\label{10} \end{align}

and we can conclude that

\begin{align} r\,R&=\frac{11}8 \tag{11}\label{11} ,\\ r^2&=-\frac52 \tag{12}\label{12} . \end{align}

With the substitution of \eqref{5}-\eqref{7}, question becomes to find

\begin{align} \Delta^2&=(b-a)^2(c-b)^2(a-c)^2 \tag{13}\label{13} . \end{align}

Using the known expression of the r.h.s of \eqref{13} in terms of $\rho,r,R$,

\begin{align} \Delta^2&= 4\,r^2\,(4R\,(\rho^2\,(5r+R)-r\,(4R^2+3\,(r+2R)^2))-(\rho^2+r^2)^2) \tag{14}\label{14} \end{align}

with $\rho=2$ we have

\begin{align} \Delta^2&= 320\,(rR)\,r^2+64\,(rR)^2-256\,(rR)^3-48\,(r^2)^2\,(rR)-192\,r^2\,(rR)^2-64\,r^2 -32\,(r2)^2-4\,(r^2)^3 . \tag{15}\label{15} \end{align}

Finally, the substitution of \eqref{11}-\eqref{12} into \eqref{15} gives the result \begin{align} \Delta^2&=-1127 \tag{16}\label{16} , \end{align}

which agrees with the other answers.

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Edit

Using the same approach, the answer for a general cubic \begin{align} x^3+u\,x^2+v\,x+w&=0 \tag{17}\label{17} \end{align}

can be found to be

\begin{align} \Delta^2&=u^2 v^2+18\,u\,v\,w-4v^3-4u^3\,w-27w^2 \tag{18}\label{18} . \end{align}