<s>[INST] Find the inverse with respect to the binary operation $a ∗ b = a + b + a^2 b^2$ A binary operation on $\mathbb{R}$:  $a * b = a + b + a^2 b^2$
The neutral element I found to be $0$. 
Then I need to find an invertible element having two distinct inverses. I don't know where to start for this question.
 [/INST] If $a = 0$, $b = 0$ is the unique inverse of $a$, since $0 = 0 * b = 0 + b + 0^2b^2 = b$ in this case.
If $a \ne 0$, $a * b = a + b + a^2b^2 = 0$ is a quadratic equation for $b$.  It may be written as
$b^2 + \dfrac{1}{a^2}b + \dfrac{1}{a} = 0. \tag{1}$
Applying the quadratic formula to (1) yields
$b = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4}{a}}), \tag{2}$
which may be simplified somewhat:
$b = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4}{a}}) = \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1}{a^4} - \dfrac{4a^3}{a^4}})$
$= \dfrac{1}{2}(-\dfrac{1}{a^2} \pm \sqrt{\dfrac{1 - 4a^3}{a^4}}) = \dfrac{1}{2a^2}(-1 \pm \sqrt{1 - 4a^3}),  \tag{5}$
which has one value, $b = (-1/2a^2)$, precisely when
$4a^3 = 1 \; \; \text{or} \; \; a = \dfrac{1}{\sqrt[3]{4}}; \tag{6}$ 
if $0 \ne 4a^3 < 1$, we obtain two real solutions for $b$, so there are two inversens for any such $a$; if $4a^3 > 1$, then the values of $b$ form a complex conjugate pair; their is no (real) inverse for such $a$.  But any nonzero $a < (1 / \sqrt[3]{4})$ will have exactly two inverses in $\Bbb R$ for the operation $a * b = a + b + a^2b^2$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
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