Derivating a polynomial equation from two others

Question

In an old article of J. Cockle, Sketch of a Theory of Transcendental Roots, pages $146$/$147$, the author obtains an expression for a cubic equation I cannot reproduce. This equation, when you solve this, results into the correct numerical values for the roots of the cubic.
The equations are: \begin{align*} x^3-3x+2a = 0\\3x^2\frac{dx}{da}-3\frac{dx}{da}+2=0 \end{align*} Eliminating $x$ leads to : $$\left(a^2-1\right)\left(3\frac{dx}{da}\right)^3+9\frac{dx}{da}+2=0$$ Can anybody derive this last expression from the first two?

Answer 1

From the second equation, you have $$\frac{dx}{da}=\frac{2}{3(1-x^2)}\implies \left(3\frac{dx}{da}\right)^3=\frac{8}{(1-x^2)^3}$$

Therefore, $$\left(a^2-1\right)\left(3\frac{dx}{da}\right)^3+9\frac{dx}{da}=\frac{8(a^2-1)}{(1-x^2)^3}+\frac{6}{1-x^2}$$

Meanwhile, from the first equation, $$8a^2=2x^2(3-x^2)^2$$

So if you substitute this into the expression above and simplify the fraction on the right hand side, you get $$\frac{2x^6-6x^4+6x^2-2}{(1-x^2)^3}=-2,$$ thus establishing the third equation

Answer 2

Elaborating on my previous hint, let $\,\dfrac{dx}{da} = t\,$ then the problem is to eliminate $\,x\,$ between:

$$ \begin{align} x^3 - 3x + 2a = 0 \\ 3t\, x^2 - 3 t + 2=0 \end{align} $$

Writing the first equation as $\,- 2 a = x(x^2 - 3)\,$, squaring, and substituting $\,x^2 = 1 - \dfrac{2}{3t}\,$ from the second equation:

$$ 4a^2 = x^2(x^2-3)^2 = \left(1 - \dfrac{2}{3t}\right)\left(-2 \left(1 + \frac{1}{3t}\right)\right)^2 = \ldots = \frac{4(27 t^3 - 9 t - 2)}{27 t^3} $$ $$ \implies\quad 27(a^2 - 1) t^3 + 9 t + 2 = 0 $$

Remembering that $\,t = \dfrac{dx}{da}\,$, the latter is OP's third equation.

Answer 3

From

$$ 3x^2\frac{dx}{da}-3\frac{dx}{da}+2=0\Rightarrow 3x^3\frac{dx}{da}-3x\frac{dx}{da}+2x=0 $$

and then

$$ 3(3x-2a)\frac{dx}{da}-3x\frac{dx}{da}+2x=0\Rightarrow 6x\frac{dx}{da}-6a\frac{dx}{da}+2x=0 $$

now eliminating $x$ between

$$ \cases{ x^3-3x+2a=0\\ 6x\frac{dx}{da}-6a\frac{dx}{da}+2x=0 } $$

we get

$$ a\left(\left(a^2-1\right) \left(3\frac{dx}{da}\right)^3+9 \frac{dx}{da}+2\right)=0 $$