Showing that $\sqrt[3]{9+9\sqrt[3]{9+9\sqrt[3]{9+\cdots}}} - \sqrt{8-\sqrt{8-\sqrt{8+\sqrt{8-\sqrt{8-\sqrt{8+\cdots}}}}}} = 1$?

took me a while. The roots of your $$ u^3 - 9u - 9 = 0 $$ are $$ -2 \sqrt 3 \sin {\frac{2\pi}{9}} \; , $$ $$ -2 \sqrt 3 \sin {\frac{8\pi}{9}} \; = \; -2 \sqrt 3 \sin {\frac{\pi}{9}} , $$ $$ -2 \sqrt 3 \sin {\frac{14\pi}{9}} \; = \; 2 \sqrt 3 \sin {\frac{4\pi}{9}} . $$

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? poldisc( s^3 - 9 * s - 9)
%51 = 729
? polgalois( s^3 - 9 * s - 9)
%59 = [3, 1, 1, "A3"]

? polroots( s^3 - 9 * s - 9)
%52 = [-2.226681596905677465811651808 ,
       -1.184792530904095372701352048 , 
        3.411474127809772838513003856 ]~


? u = -2 * sqrt(3) * sin (2 * Pi / 9  )
%53 = -2.226681596905677465811651808
? v = -2 * sqrt(3) * sin (8 * Pi / 9  )
%54 = -1.184792530904095372701352048
? w = -2 * sqrt(3) * sin (14 * Pi / 9  )
%55 = 3.411474127809772838513003856
? 
? u^3 - 9 * u - 9
%56 = 0.E-27
? v^3 - 9 * v - 9
%57 = 0.E-27
? w^3 - 9 * w - 9
%58 = -8.07793567 E-28
?

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You also asked about $$x =\sqrt{ 5+ \sqrt{ 5 + \sqrt{ 5-x}}}$$ I figured out how to express $x$ in radicals.

Take $$ k = \frac{1 + \sqrt {13}}{2} $$ so that $$ k^2 - k + 2 = 5 \; . $$

Then $x$ above is the largest root of the cubic $$ x^3 + (k-1) x^2 - (k^2 + 2)x - (k^3 - k^2 + 2 k - 1) $$ which is the same as $$ x^3 + (k-1) x^2 - (k+5)x - ( 5 k - 1) $$

? k = (1 + sqrt(13))/2
%5 = 2.302775637731994646559610634
?  h = x^3 + (k-1) * x^2 - (k^2 + 2) * x - (k^3 - k^2 + 2 * k - 1 )
%6 = x^3 + 1.302775637731994646559610634*x^2 
    - 7.302775637731994646559610634*x - 10.51387818865997323279805317
? polroots(h)
%7 = [-2.549662363373543544157989866 + 0.E-28*I,
      -1.500778167203563600102045353 + 0.E-28*I,
       2.747664892845112497700424585 + 0.E-28*I]~
? w = x^3 +(k-1) * x^2 - (k+5) * x - (5 * k - 1) 
%8 = x^3 + 1.302775637731994646559610634*x^2 - 7.302775637731994646559610634*x 
       - 10.51387818865997323279805317
? polroots(w)
%9 = [-2.549662363373543544157989866 + 0.E-28*I, 
      -1.500778167203563600102045353 + 0.E-28*I, 
       2.747664892845112497700424585 + 0.E-28*I]~
?

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In general, with positive integer $a \geq 2$ and real $k > 0$ such that $$ k^2 - k + 2 = a, $$ so that $$ k = \frac{1 + \sqrt{4a-7}}{2} \; , $$ with $$x =\sqrt{ a+ \sqrt{ a + \sqrt{ a-x}}},$$ we find $x$ is the largest root of $$ x^3 + (k-1) x^2 - (a+k)x - ( a k - 1) \; .$$ We can use Cardano at this point. The bad news is that the cube roots will be of complex numbers, we are in Casus Irreducibilis.

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