I'll ask this question by giving a concrete example, which in my opinion is easier to understand:
Premise: it is assumed that it is possible to do this because the two equations are connected related
Easy example
By hypothesis we know that the plastic constant $\rho$ solve this equation: $x^3-x-1=0$
And the following equation is given:
$27x^{3}+9x^{2}-3x+7=0$
We need to express the real root of this equation in terms of the root of the first:
The real solution to this equation is $x=-\dfrac{2}{3\rho}-\dfrac{1}{3}=-\dfrac{2}{3}\rho^{2}+\dfrac{1}{3}$
This case is easy because you just need to manipulate the $x$ with sum and product operations.
But if for example I have an equation like this:
$$91125 x^6−425250 x^5+888975 x^4−607500 x^3+53955 x^2−690 x+3481=0$$
The real solutions are $x_1\approx 0.2565$ and $x_2\approx 0.9836$
I know that both are expressible in terms of $\rho$, but I don't know how to calculate it.
The only consideration I was able to make was consider $3\sqrt{5}x=t$ and simplify the equation to
$$t^{6}-14\sqrt{5}t^{5}+439t^{4}-900\sqrt{5}t^{3}+1199t^{2}-46\sqrt{5}t+3481=0$$
Probably this equation can be divided into a product of this type: $$\left(t^3+(a_2+b_2)t^2+(a_1+b_1)t+(a_0+b_0)\right)\left(t^3+(a_2-b_2)t^2+(a_1-b_1)t+(a_0-b_0)\right)=0$$ But it seems like a very long process to me, taking into account that I would also have to do this with higher degree polynomials.
Are there any faster methods that only require you to manipulate the "secondary" polynomial with the "primary" polynomial?
Update
The polynomial is the product of
$$p_1(x)=135\sqrt{5}x^{3}-45(7\sqrt{5}\color{red}{+}10)x^{2}+3(147\sqrt{5}\color{red}{+}340)x-101\sqrt{5}\color{red}{-}218$$ and $$p_2(x)=135\sqrt{5}x^{3}-45(7\sqrt{5}\color{red}{-}10)x^{2}+3(147\sqrt{5}\color{red}{-}340)x-101\sqrt{5}\color{red}{+}218$$
$x_1$ solves $p_1(x)$ and $x_2$ solve $p_2(x)$
Maybe you can use the same idea as with the plastic constant which is also the sum of two real roots.
\begin{align} r_1 = & \left(\frac{1}{2}+\frac{1}{6}\sqrt{\frac{23}{3}}\right)^{1/3} \qquad r_2 = \left(\frac{1}{2}-\frac{1}{6}\sqrt{\frac{23}{3}}\right)^{1/3} \\ \rho =& \;r_1 +r_2 = \frac{(9-\sqrt{69})^{1/3}+(9+\sqrt{69})^{1/3}}{2^{1/3} 3^{2/3}} = 1.324717957.. \end{align}
Notice that the quadratic polynomial combining $r_1,r_2$ has simpler roots : $$ x^2-\frac{1}{9} = (x- r_1 r_2) (x+r_1 r_2) = \left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)$$
For the first polynomial $27x^{3}+9x^{2}-3x+7=0$ with only one real root you solved the relationsship without adding a new constant.
If for the second polynomial $$91125 x^6−425250 x^5+888975 x^4−607500 x^3+53955 x^2−690 x+3481=0$$ one uses the same idea as with the plastic constant one obtains a quadratic polynomial with roots which are of course more complicated but they also have $\sqrt{69}$ like the plastic constant as you can see:
$$x^2-\frac{1}{15^4}\left(-35 + \frac{\left(2^5\cdot 15^3\cdot 37 - 2^5\cdot 3\cdot 5^4\cdot 7 \sqrt{69}\right)^{1/3}}{3} + 10 \left(\frac{2}{3}\right)^{2/3} \left(3^2\cdot 7 + 35 \sqrt{69}\right)^{1/3}\right)^2 $$