Prove that a property holds for the roots of an equation, given some constraints. For example, in this question:

The cubic expression $$ ax^3 + bx^2 + cx + d $$ has a pair of roots which are reciprocals of each other. Prove that $$a^2-d^2=ac-bd$$.

I understand that the correct approach is probably to use the formulae relating coefficients and roots, but I'm not exactly sure where to go for there.
Is there are a more generic framework for solving these types of problems?
 A: Let the roots be $ y, z, \frac{1}{z}$.
As you said, "use the formulae relating coefficients and roots" to show that

*

*$y+z + \frac{1}{z} = -b/a$

*$yz + y/z + 1 = c/a$

*$y = -d/a$
Finally, verify that
$$1 - (d/a)^2 = 1-y^2 = (c/a) - (b/a)(d/a).$$
Hence, $ a^2 - d^2 = ac - bd$.
A: As a complement to @CalvinLin's answer, let us denote the polynomials
\begin{equation}
\begin{array}[l]\cr
P(x) = a x^3 + b x^2 + c x + d\cr
Q(x) = d x^3 + c x^2 + b x + a
\end{array}
\end{equation}
Explicit computation shows that the resultant of these two polynomials is
\begin{equation}
R(P, Q) = \left|\begin{matrix}
a&0&0&d&0&0\cr
b&a&0&c&d&0\cr
c&b&a&b&c&d\cr
d&c&b&a&b&c\cr
0&d&c&0&a&b\cr
0&0&d&0&0&a\cr
\end{matrix}\right| = -(a^2-d^2- a c + b d)^2 P(1)P(-1)
\end{equation}
It follows that $P$ and $Q$ have a common root if and only if $1$ or $-1$ is a root of $P$ or $a^2-d^2 = a c - b d$. Note that $1$ and $-1$ are their own reciprocals.
Here is the (python) code to prove the claim
import sympy as sp

a, b, c, d = sp.symbols(['a','b','c','d'])

M = sp.Matrix([
    [a, b, c, d, 0, 0],
    [0, a, b, c, d, 0],
    [0, 0, a, b, c, d],
    [d, c, b, a, 0, 0],
    [0, d, c, b, a, 0],
    [0, 0, d, c, b, a],
])
p = a**2 - d**2 - a * c + b * d
q = (a + b + c + d) * (a - b + c - d)
R = M.det()
s = R - p**2 * q
assert s.simplify() == 0

A: Okay, so if $r$ and $\frac 1r$ are two roots of $ax^3 + bx^2 + cx + d$ then $ax^3 + bx^2 + cx + d = a(x-r)(x-\frac 1r)(x+k)$ for some $k$.
So $ka = d$
And $ak -ar -a\frac 1r = d-ar -a\frac 1r = b$.
$-akr-ak\frac 1r+a = -dr-d\frac 1r + a=c$
If we assume $a\ne 0$ then $r+\frac 1r = \frac {d-b}a$ and $a = c+\frac {d-b}a\cdot d$.
So $a^2 =ca +d(d-b) = ca + d^2 -bd$ and $a^2 -d^2 =ca -db$.
And if $a = 0$ then $d=b$. and the result follows.
(If $a = 0; b\ne 0$ then $bx^2 + cx +d = b(x-r)(x-\frac 1r)= bx^2 -b(r+\frac 1r)x + b$ so $b=d$ )
