Is there an analytic solution to this set of cubic equations? I have three parameters $(a,b,c)$ that define equations I am looking for an analytical solution.
$$q_1=b/c$$
$$q_2=\frac{(a+b)^3}{3c^2b}$$
$$q_3=\frac{(a+b)^3}{3b^3}$$
I want to solve these equations for $a$, $b$, and $c$ in terms of $q_1$, $q_2$, and $q_3$. The first equation can easily be solved for $b=q_1c$. Inserting this into either of the two remaining is third degree of each parameter.
$$q_2=\frac{a^3}{3c^3q_1}+\frac{a^2}{c^2}+\frac{aq_1}{c}+\frac{q_1^2}{3}$$
$$q_3=\frac{a^3}{27c^3q_1^3}+\frac{a^2}{9c^2q_1^2}+\frac{a}{cq_1}+\frac{1}{3}$$
This doesn't seem like a problem where I could apply the cubic formula, unless there is no error in defining a new dummy variable $x=a/c$ first and solving for its constituents later.
If there isn't an "easy" analytical solution to find I'll linearize the last two equations, but I'm looking for something more exact.

Slightly rewriting the equations to avoid fractions:
$$q_1 c = b \tag{1}$$
$$3 q_2 bc^2 = (a + b)^3 \tag{2}$$
$$3 q_3 b^3 = (a + b)^3 \tag{3}$$
From (2) and (3), we get $3 q_2 bc^2 = 3 q_3 b^3$, or $q_2 c^2 = q_3 b^2$.  But from (1), $b = q_1 c$, so plugging that in gives $q_2c^2 = q_3 q_1^2 c^2$, or $q_2 = q_1^2 q_3$.  So we don't have three independent $q$'s.  I'll deal with this redundancy later.
Taking the cube root of (3) gives:
$$b\sqrt[3]{3q_3} = a + b$$
Or equivalently, $a = (\sqrt[3]{3q_3} - 1)b$.
From (1), we get $c = \frac{b}{q_1}$.  And with explicit formulas for $a$ and $c$ in terms of $b$, we can now write (2) in terms of $b$ alone.
$$3 q_2 b(\frac{b}{q_1})^2 = 3q_3 b$$
Cancelling $b$ from each side gives an easily-solved quadratic, with $b = q_1 \frac{\sqrt{q_3}}{\sqrt{q_2}}$.
Thus, the solution is:
$$a = (\sqrt[3]{3q_3} - 1)q_1\frac{\sqrt{q_3}}{\sqrt{q_2}}$$
$$b = q_1 \frac{\sqrt{q_3}}{\sqrt{q_2}}$$
$$c = \frac{\sqrt{q_3}}{\sqrt{q_2}}$$
However, as noted earlier, the three $q$'s aren't independent, but related by $q_2 = q_1^2 q_3$.  So we only need two of these to express our solution.
If we substitute $q_1 = \frac{\sqrt{q_2}}{\sqrt{q_3}}$, then:
$$a = (\sqrt[3]{3q_3} - 1)$$
$$b = 1$$
$$c = \frac{\sqrt{q_3}}{\sqrt{q_2}}$$
If we substitute $q_2 = q_1^2 q_3$, then:
$$a = (\sqrt[3]{3q_3} - 1)$$
$$b = 1$$
$$c = \frac{1}{q_1}$$
Finally, if we substitute $q_3 = \frac{q_2}{q_1^2}$,
$$a = (\sqrt[3]{3\frac{q_2}{q_1^2}} - 1)$$
$$b = 1$$
$$c = \frac{1}{q_1}$$
So $b$ isn't really a variable at all!
