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.

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.
