# How to solve the following system : $\begin{cases} \ \ \ ab^2 = 2 \\ a+b = -1 \end{cases}$?

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{aligned} a(-1-a)^2=2&\Leftrightarrow a(a^2+2a+1)=2 \\ &\Leftrightarrow a^3+2a^2+a-2=0 \\ \end{aligned}
• There is an error in the second line: it should be $+2a^2$, not minus. – Ennar Apr 29 '18 at 11:51
• I have edited, however the solution will be quite ugly: $a=0.6956207696; b=-1.6956207696$. – user061703 Apr 29 '18 at 11:55
• @Jam For anyone who can't access to the link: $a=\frac{1}{3}(-2+\sqrt[3]{28-3\sqrt{87}}+\sqrt[3]{28+3\sqrt{87}})$ – user061703 Apr 29 '18 at 11:58