Solving $\beta^3-2\beta^2+(1-\rho)=0$ for $\beta$

Asked Apr 20 '20 at 13:03

Active Apr 20 '20 at 16:53

Viewed 89 times

I need to solve the following equation for $\beta$: $$\beta^3-2\beta^2+(1-\rho)=0$$ where $\rho$ is just a constant.

I already tried different kind of methods from the wikipedia page for cubic equations and various other site without luck.

The answer should be: $$\frac{2}{3}-\frac{4}{3}\sin(\frac{1}{3}\arcsin(\frac{27}{16}|\rho|-\frac{11}{16}))$$

I hope someone knows how to handle this equation.

edited Apr 20 '20 at 16:53

user26857

asked Apr 20 '20 at 13:03

esmo

Just plug into the equation and check if you get the equality. – Kenta S Apr 20 '20 at 13:05
I am not supposed to do that. I need to find this solution. – esmo Apr 20 '20 at 13:06
2

There are several ways to solve it, see for example here. You can apply a substitution to get rid of $-2\beta^2$. – Dietrich Burde Apr 20 '20 at 13:25
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See also https://en.wikibooks.org/wiki/Trigonometry/The_solution_of_cubic_equations – lhf Apr 20 '20 at 13:33
See also https://math.stackexchange.com/questions/1908861/using-trig-identity-to-solve-a-cubic-equation – lhf Apr 20 '20 at 13:43

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