Absolute Value Equation $$x^3+|x| = 0$$
One solution is $0.$ We have to find the other solution (i.e, $-1$)
$$Solution:$$
CASE $1$:
If $x<0,~|x| = -x$, we can write $x^3+|x| = 0$ as $-x^3-x=0$
$$x^3+x=0$$
$$x(x^2+1)=0$$
$$\Longrightarrow x=0, or, x=\sqrt{-1}$$
Please tell me where I've gone wrong.

the function $$f(x) = \begin{cases} x^3 - x & if \, x \le 0,\\x^3+x &if \, 0 \le x. \end{cases}  $$
has one negative zero at $x = -1$ and a zero at $x = 0.$ you can see from the function definition that $f(x) > 0$ for $x > 0.$  the graph of $y = f(x)$ has a cusp at $(0,0)$ which is also a local minimum.
