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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?

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    $\begingroup$ You might want to look at Vieta's formulae/relations which relate the coefficients of a polynomial to the elementary symmetric functions of the roots - but for this kind of problem you are probably looking for a short cut. $\endgroup$ – Mark Bennet Jan 24 at 20:35
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    $\begingroup$ Sorry, I meant reciprocals. $\endgroup$ – hidden-shelter Jan 24 at 20:37
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    $\begingroup$ I've updated it accordingly. $\endgroup$ – hidden-shelter Jan 24 at 20:37
  • $\begingroup$ Thank you for clarifying. $\endgroup$ – Mark Bennet Jan 24 at 20:38
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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$.

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  • $\begingroup$ Thanks! I'm just not sure as to how the verification leads to the proof that $a^2-d^2 = ac - bd$? $\endgroup$ – hidden-shelter Jan 24 at 20:53
  • $\begingroup$ As I stated/implied, show that $ 1 - (d/a) ^2 = 1- y^2$, and that $ 1- y^2 = (c/a) - (b/a)(d/a)$, then multiply throughout by $a^2$. $\endgroup$ – Calvin Lin Jan 24 at 20:54
  • $\begingroup$ But how does this relate to $a^2 - d^2$? $\endgroup$ – hidden-shelter Jan 24 at 20:56
  • $\begingroup$ Oh wait, no I see it now. Many thanks! $\endgroup$ – hidden-shelter Jan 24 at 20:57
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    $\begingroup$ What is $ a^2 \times [ 1 - (d/a) ^2 ]$? What is $ a^2 \times [ (c/a) - (b/a)(d/a) ] $? If you're stuck, please show what you've tried and why you're stuck. $\endgroup$ – Calvin Lin Jan 24 at 20:57
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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
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  • $\begingroup$ Why not sp.factor(R)? $\endgroup$ – Marc Glisse Jan 25 at 7:58
  • $\begingroup$ @MarcGlisse You are right. There is even a "resultant" in sympy. I should do more formal calculations... $\endgroup$ – Gribouillis Jan 26 at 15:26
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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 $b=d$ )

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