Real numbers $a$ and $b$ satisfy
a + ab^2 = 250ab + ab^3
a - ab^2 = 240ab - 50b^2
Enter all possible values of a, separated by commas.
From the given equations, we can simplify them by factoring out a common factor of $ab$:
$ab(1+b^2) = 250ab + ab^3$
$ab(1-b^2) = 240ab - 50b^2$
Dividing both sides of both equations by $ab$, we get:
$1 + b^2 = 250 + b^3$, and
$1 - b^2 = 240 - 50b^2$
The first equation simplifies to $b^3 - b^2 + 249 = 0$, and the second equation simplifies to $49b^2 - b + 239 = 0$.
By the Rational Root Theorem, the only possible rational roots of the first equation are $\pm1, \pm3, \pm83, \pm249$. Checking, none of these are roots of the equation.
By the Quadratic Formula, the possible values of $b$ that are roots of the second equation are $\frac{1 \pm \sqrt{1 - 4(49)(239)}}{2(49)}$. However, the discriminant, $1 - 4(49)(239)$, is negative, so there are no real solutions for $b$.
Therefore, there are no possible real values for $a$ and $b$ that satisfy both given equations.