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.
 A: 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!
