نه تنها خدا کمک نمی کند، بلکه خداوند شما را در تنگنا قرار نمی دهد. // Not only God does not help, but also Allah does not "put you in trouble"

          I have solved such a problem before, here.

          It's Problem #5 on page 256 of the book НЕСТРЕНКО Ю.[рий] В.[алентинович] ТЕОРИЯ ЧИСЕЛ

(Издательский центр "АКАДЕМИЯ", Москва, 2008)

In translation :

" Let $\alpha$ be a root of the polynomial $f(x)=x^3-9x-9$ and $\beta=6+\alpha-\alpha ^2\;,\;\gamma=-6-2\alpha+\alpha ^2.$ Prove that the numbers $\beta\;,\;\gamma$, are also roots of the polynomial $f(x)$."

As we have seen before, the main difficulty of the problem lies in writing certain expressions as perfect squares.

Solution CiP

Obvious

$$\alpha ^3=9\alpha+9. \tag{1}$$

From here we get, first multiplying (1) by $\alpha$

$$\alpha ^4=9\alpha ^2+9\alpha\;, \tag{2}$$

then, from $\alpha \cdot (\alpha^2-9)=9$, that

$$\frac {1}{\alpha}=\frac{1}{9}\cdot \alpha ^2-1. \tag{3}$$

From the first formula of Vieta $\alpha +\beta +\gamma=0$ we get

$$\beta+\gamma=-\alpha, \tag{4}$$

and from the third $\alpha \cdot \beta \cdot \gamma=9$ we deduce $\beta \cdot \gamma=\frac{9}{\alpha}$  therefore (see relation (3))

$$\beta \cdot \gamma=\alpha^2-9. \tag {5}$$

We calculate now, $(\beta -\gamma)^2=(\beta+\gamma)^2-4\cdot \beta \cdot \gamma \underset {(4)\;(5)}{==}$

$$=(-\alpha)^2-4(\alpha^2-9)=36-3\alpha^2. \tag{6}$$

Or we can, from $\beta \cdot \gamma=\frac{9}{\alpha}$, calculate the same way

$(\beta-\gamma)^2=(\beta+\gamma)^2-4\cdot \beta \gamma=(-\alpha)^2-4 \cdot \frac{9}{\alpha}=\frac{\alpha^3-36}{\alpha}\underset{(1)}{=} \frac{9\alpha+9-36}{\alpha}=$

$$=9 \cdot \frac{\alpha -3}{\alpha}=9\cdot \frac{\alpha^2-3\alpha}{\alpha^2}. \tag{7}$$

          It would be preferable to have perfect squares in relations (6) and/or (7). The method of undetermined coefficients did not help me this time. I don't know by what miracle I finally determined (which can be verified immediately by calculation) the following:

$36-3\alpha^2=(2\alpha^2-3\alpha-12)^2, $ and

$\alpha^2-3\alpha=(\alpha^2-2\alpha-6)^2.$

Now, from (6) it follows $\beta-\gamma=12+3\alpha-2\alpha^2$ (making one of the possible choices of $\pm$ signs) which combined with (4) leads to the desired values

 $$\beta=6+\alpha -\alpha^2\;, \;\gamma=-6-2\alpha+\alpha^2.$$

$\blacksquare$