'Elegant' ways on solving for roots for this cubic function?

Question

I have this interesting cubic equation,

$$ x^{3} - 80\alpha x^{2} + (1744\alpha^{2}-81)x + (3240\alpha-5760\alpha^{3}) = 0 $$

where $\alpha$ is some constant.

I went about the method of Cardano, Tartaglia, and del Ferro, and so far with pages of work, all these substitutes have become too damned mind boggling.

Is their a much more better way of solving this without making about 5-7 substitutions?

Answer 1

Hint: $x = 40\alpha$ is one root of this equation.

This is the way I found it.

First, I would expect this type of equation has a "nice" root.

Second, if I consider $\alpha$ as a variable, then the LHS can be written as sum of a polynominal of degree $3$ and a polynominal of degree $1$, where the polynominal of degree $1$ is $-81x+3240\alpha$.

Third, I would expect each of these polynominals is $0$, which gives me $x = 40 \alpha$ in the second one. Plug this in the first one, it also gives $0$.

Answer 2

According to Wolfy, the roots are $40a$ and $x = 20 a\pm\sqrt{256 a^2+81} $.

The factorization is $(x-40 a) (144 a^2-40 a x+x^2-81) = 0 $.