Solving for $a$ and $b$

Question

How would you solve the equality: $$a\,b\,\left(b+a\right)=1$$ in terms of a and/or b?
Would you subtract 1 from both sides and work from there? Or would you simply expand and work from there?

Answer 1

For the sake of closure I'll "answer" Pedro's answer in the comments. we have $ab^2+a^2b-1=0$, so we can use the quadratic equation to solve for one in terms of the other. This yields

$$b=\frac{-a^\frac{3}{2}\pm\sqrt{a^2+4}}{2\sqrt{a}}$$

Answer 2

Let $b=ta$ for $t\notin\{-1,0\}$. Then $1=a^3t(t+1)$, that is $a=1/\sqrt[3]{t(t+t)}$ and $b=t/\sqrt[3]{t(t+1)}$.