Solving $a = b^2 + 2b^2(1 - b)$ for $b$ My algebra is very rusty, it's been about 15years since I studied, and I was stumped recently when trying to rearrange this formula;
$$a = b^2 + 2b^2(1 - b)$$
to give $b$ in terms of $a$.
Can someone show me step by step working please :)
I remember 'change side, change sign' but it all gets very confusing very quickly.
Thanks!

So, 
$$2b^2(1-b)\Rightarrow 2b^2 - 2b^3$$
Giving
$$a = b^2 + 2b^2 - 2b^3$$
$$a = 3b^2 - 2b^3$$
$$a + 2b^3 - 3b^2 = 0$$
is this right so far?

Your last equation is correct.
$$\begin{equation*}
a+2b^{3}-3b^{2}=0
\end{equation*}\tag{0}$$
This equation is a cubic equation in $b$, which can be solved algebraically by the Cardano's cubic formula (see e.g. this PlanetMath entry). We can write it as
$$\begin{equation*}
b^{3}-\frac{3}{2}b^{2}+\frac{a}{2}=0.
\end{equation*}\tag{1}$$

*

*This equation has in general $3$ complex roots. First we make the change of variables $$b=t+\frac{1}{2}.\tag{1a}$$ Since $$\begin{equation*}
\left( t+\frac{1}{2}\right) ^{3}-\frac{3}{2}\left( t+\frac{1}{2}\right) ^{2}+\frac{a}{2}=t^{3}-\frac{3}{4}t-\frac{1}{4}+\frac{1}{2}a,
\end{equation*}\tag{2}$$ we get the reduced cubic equation $$\begin{equation*}
t^{3}-\frac{3}{4}t-\frac{1}{4}+\frac{1}{2}a=0.\tag{3}
\end{equation*}$$

*The Cardano's method to solve $(3)$ is to write the variable $t$  as a sum of two variables $$t=u+v.\tag{3a}$$ Since
$$\begin{eqnarray*}
&&\left( u+v\right) ^{3}-\frac{3}{4}\left( u+v\right) -\frac{1}{4}+\frac{1}{2}a \\
&=&\left( u^{3}+v^{3}-\frac{1}{4}+\frac{1}{2}a\right) +\left( 3uv-\frac{3}{4}\right) \left( u+v\right) =0
\end{eqnarray*}\tag{4}$$
every solution of the following system is a solution of the reduced cubic $(3)$ as well. $$
\begin{eqnarray*}
&&\left\{ 
\begin{array}{c}
u^{3}+v^{3}-\frac{1}{4}+\frac{1}{2}a=0 \\ 
3uv-\frac{3}{4}=0
\end{array}
\right.  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
u^{3}+v^{3}=\frac{1}{4}-\frac{1}{2}a \\ 
u^{3}v^{3}=\frac{1}{64}
\end{array}
\right.  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
u^{3}+\frac{1}{64u^{3}}-\left( \frac{1}{4}-\frac{1}{2}a\right) =0 \\ 
v^{3}=\frac{1}{64u^{3}}
\end{array}
\right.  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
\left( u^{3}\right) ^{2}-\left( \frac{1}{4}-\frac{1}{2}a\right) u^{3}+\frac{1
}{64}=0 \\ 
v^{3}=\frac{1}{64u^{3}}.
\end{array}
\right. 
\end{eqnarray*}\tag{5}$$

*If we write $$Y=u^{3},\tag{5a}$$ we get the following quadratic equation
$$\begin{equation*}
Y^{2}-\left( \frac{1}{4}-\frac{1}{2}a\right) Y+\frac{1}{64}=0,
\end{equation*}\tag{6}$$ whose solutions are $$\begin{eqnarray*}
Y_{+} &=&\frac{1}{8}-\frac{1}{4}a+\frac{1}{4}\sqrt{-a+a^{2}} \\
Y_{-} &=&\frac{1}{8}-\frac{1}{4}a-\frac{1}{4}\sqrt{-a+a^{2}}.
\end{eqnarray*}\tag{7}$$

*So, without loss of generality, the solutions of the system $(5)$ are
$$\begin{equation*}
u=Y_{+}^{1/3},\qquad v=\frac{1}{4Y_{+}^{1/3}}=Y_{-}^{1/3}.
\end{equation*}\tag{8}$$

*Combining the above results we obtain the following formula: $$\begin{eqnarray*}
b &=&t+\frac{1}{2}=u+v+\frac{1}{2} \\
&=&\left( \frac{1}{8}-\frac{1}{4}a+\frac{1}{4}\sqrt{a^{2}-a}\right)
^{1/3}+\left( \frac{1}{8}-\frac{1}{4}a-\frac{1}{4}\sqrt{a^{2}-a}\right)
^{1/3}+\frac{1}{2}.\tag{9}
\end{eqnarray*}$$

*The $3$ complex (or real) values of $(9)$ give all the solutions of $(0)$.

