Solving Nonlinear System: \alpha,\beta,\gamma from A,B,C

[Bruno Tolentino]

I have this system of equation: $$A = \frac{\alpha + \beta + \gamma}{3}$$ $$B = \sqrt[2]{\frac{\beta \gamma + \gamma \alpha + \alpha \beta}{3}}$$ $$C = \sqrt[3]{\alpha \beta \gamma}$$ And I want to solve this system for α, β and γ. In other words, I want to express α, β and γ in terms of A, B and C.

$$\alpha = \alpha (A,B,C)$$$$\beta = \beta (A,B,C)$$$$\gamma = \gamma (A,B,C)$$

 

[mfb]

What is your question or discussion point?
There is an obvious way to start solving it.

 

[Bruno Tolentino]

If α and β are the roots of the quadratic equation and A and B are the arithmetic and geometric mean, respectively, so, the quadratic formula becomes: $$A \pm \sqrt{A^2-B^2}$$. I'm trying to solve the cubic equation in the same way...

 

[mfb]

I would introduce new variables for B² and C³. Solving the third equation for one variable and plugging it into another equation gives a quadratic equation for a second variable, which can be plugged into the third equation. If that has a degree of at most 4, there is an analytic solution in closed form, otherwise I doubt there is a way to solve it (because a solution would then probably allow to solve equations that are proven to have no closed analytic solution).

 

[HallsofIvy]

I have this system of equation: $$A = \frac{\alpha + \beta + \gamma}{3}$$ $$B = \sqrt[2]{\frac{\beta \gamma + \gamma \alpha + \alpha \beta}{3}}$$ $$C = \sqrt[3]{\alpha \beta \gamma}$$ And I want to solve this system for α, β and γ. In other words, I want to express α, β and γ in terms of A, B and C.

$$\alpha = \alpha (A,B,C)$$$$\beta = \beta (A,B,C)$$$$\gamma = \gamma (A,B,C)$$

So, essentially, you want to solve
$$\alpha+ \beta+ \gamma= 3A$$
$$\alpha\beta+ \alpha\gamma+ \beta\gamma= 9B^2$$
$$\alpha\beta\gamma= C^3$$
and since A, B, and C are given values, so are 3A, $9B^2$, and $C^3$.

From $\alpha\beta\gamma=C^3$, $\gamma= \frac{C^3}{\alpha\beta}$
so $\alpha\beta+ \frac{C^3}{\beta}+ \frac{C^3}{\alpha}= 9B^2$
Multiplying by $\alpha\beta$, $\alpha^2\beta^2+ C^3\alpha+ C^3\beta= 9B^2\alpha\beta$.

We can write that as $\alpha^2\beta^2+ C^3\beta+ (C^3\alpha- 9B^2\alpha)= 0$ and use the quadratic formula to solve for $\beta$ in terms of $\alpha$, then put that into $\gamma= \frac{C^3}{\alpha\beta}$ to get $\gamma$ in terms of $\alpha$ only.

Finally, put those into $\alpha+ \beta+ \gamma= 3A$ to get an equation in $\alpha$ only.