Could you explain to me please, how to solve the following system of $ 2 $ equations in $ \mathbb{R} $ or $ \mathbb{C} $ : $$ \begin{cases} \ \ \ ab^2 = 2 \\ a+b = -1 \end{cases} $$ According to my opinion, this system seems to correspond to a system in the form : $$ \begin{cases} \ \ \ ab = 2 \\ a+b = -1 \end{cases} $$ which we can solve by solving the following equation of second degree : $ x^2 + x + 2 = 0 $
But here, there is a number in exponent of $ b$ : $ 2 $ in $ ab^2 = 2 $, which doesn't help me to solve the system. Can you help me please ?
Thank you.
Replace $b$ with $-1-a$ in the first equation (because $a+b=-1$ is given), we will have
\begin{equation}\begin{aligned} a(-1-a)^2=2&\Leftrightarrow a(a^2+2a+1)=2 \\ &\Leftrightarrow a^3+2a^2+a-2=0 \\ \end{aligned}\end{equation}
You need to solve a cubic equation of third degree, not second degree.