Question 155
Find all real numbers $a$ for which there are different real numbers $x, y, z$ such that
$$
\frac{x^3 + a}{y+z} = \frac{y^3 + a}{x+z} = \frac{z^3 + a}{x+y} = -3.
$$
$$
\frac{x^3 + a}{y+z} = \frac{y^3 + a}{x+z} = \frac{z^3 + a}{x+y} = -3.
$$