Let $r,s,t$ are roots of the cubic equation $x^3+bx^2+cx+d=f(x)$ then write down $D=((r-s)(r-t)(s-t))^2$ in terms of $b, c, d$ Let $r,s,t$ are roots of the cubic equation $f(x) = x^3+bx^2+cx+d$ then write down $D=((r-s)(r-t)(s-t))^2$ in terms of $b,c,d$.
Is there any clever way to solve it?
We know $r+s+t=-b, rs+st+rt=c, rst=d.$ I also have the feeling that $D$ is actually defined as a discriminant of the cubic so there must be a discriminant of a matrix or some tricky way to prove that this is in particular ${\displaystyle b^{2}c^{2}-4c^{3}-4b^{3}d-27d^{2}+18bcd\,.}$ I am just not getting the answer without making my hands dirty i.e. expanding both sides.
 A: Let
$$M = \begin{bmatrix}
1 & 1 & 1\\
r & s & t\\
r^2 & s^2 & t^2
\end{bmatrix}.
$$
$M$ is a Vandermonde matrix, so $\det(M) = \prod_{\mathrm{cyc}} (r - s)$ and $\det(MM^\top) = D = \prod_{\mathrm{cyc}}(r - s)^2$. Thus if we define $\sigma_n = r^n + s^n + t^n$ for every $n \geq 0$, then one can easily compute
$$
MM^\top = \begin{bmatrix}
\sigma_0 & \sigma_1 & \sigma_2\\
\sigma_1 & \sigma_2 & \sigma_3\\
\sigma_2 & \sigma_3 & \sigma_4
\end{bmatrix}.
$$
Now you only need to calculate $\sigma_i$ for $0 \leq i \leq 4$ and $\det(MM^\top)$.

It's unavoidable to do some calculations for getting an expression like $b^2c^2 - 4c^3 - 4b^3d - 27d^2 + 18bcd$. But I think that one can simply understand this solution and also, generalize it for higher degree polynomials.
A: Your approach is correct. It is the discriminant of the cubic. But there is no reason to reinvent the wheel. We follow the standard procedure. We produce the depressed cubic by setting $y=x+\dfrac{b}{3}$ which is $y^{3}+py+q=0$ where $p=\dfrac{3c-b^{2}}{3}$ and $q=\dfrac{2b^{3}-9bc+27d}{27}$ and you prove that the discriminant of the depressed cubic is
$-(4p^{3}+27q^{2})$ and setting the above values we get
$D=18bcd-4b^{3}d+b^{2}c^{2}-4c^{3}-27d^{2}$
A: Remark: A solution using middle school skill.
We have
$$(r-s)(r-t)(s-t) = r^2s + s^2t + t^2r - r^2t - t^2s - s^2 r = A - B$$
where
$$A := r^2s + s^2t + t^2r, \quad B := r^2t + t^2s + s^2 r.$$
We have
$$[(r-s)(r-t)(s-t)]^2 = (A - B)^2 = (A + B)^2 - 4AB.$$
We have
\begin{align*}
 A + B &= r^2(s + t) + t^2(r + s) + s^2(t + r)\\
 &= (r^2 + t^2 + s^2)(s + t + r) - (r^3 + t^3 + s^3).
\end{align*}
We have
$$
 AB = rst(r^3 + t^3 + s^3) + (r^3s^3 + s^3t^3 + t^3r^3) + 3r^2s^2t^2.
$$
We have $r + s + t = -b, ~ rs + st + tr = c, ~ rst = -d$.
We have
$$r^2 + t^2 + s^2 = (r + s + t)^2 - 2(rs + st + tr) = b^2 - 2c. \tag{1}$$
Using the known identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$, we have
\begin{align*}
 r^3 + t^3 + s^3 &= 3rst + (r + s + t)(r^2 + t^2 + s^2 - rs - st - tr)\\
 &= -3d - b (b^2 - 3c)\\
 &= -3d - b^3 + 3bc. \tag{2}
\end{align*}
Using the known identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$, we have
\begin{align*}
 &r^3s^3 + s^3t^3 + t^3r^3\\
 =\,& 3rs\cdot st \cdot tr + (rs + st + tr)[(rs)^2 + (st)^2 + (tr)^2 - rs\cdot st - st\cdot tr - tr\cdot rs]\\
 =\,&3(rst)^2 + (rs + st + tr)[(rs + st + tr)^2 - 3rst(r + s + t)]\\
 =\,& 3d^2 + c(c^2 - 3bd)\\
 =\,& 3d^2 + c^3 - 3bcd. \tag{3}
\end{align*}
Using (1)-(3), we have
$$A + B = (b^2 - 2c)(-b) - (-3d - b^3 + 3bc)
= 3d - bc$$
and
$$AB = -d(-3d - b^3 + 3bc) + (3d^2 + c^3 - 3bcd) + 3d^2 = 9d^2 + db^3 - 6bcd + c^3. $$
Thus, we have
\begin{align*}
 (A + B)^2 - 4AB &= (3d - bc)^2 - 4(9d^2 + db^3 - 6bcd + c^3)\\
 &= -4b^3d + b^2c^2 + 18bcd - 4c^3 - 27d^2 
\end{align*}
which gives
$$[(r-s)(r-t)(s-t)]^2 = -4b^3d + b^2c^2 + 18bcd - 4c^3 - 27d^2.$$
We are done.
