Faster way to solve a equation. Solve the equation:
$\sqrt[3]{x-2} + \sqrt[3]{x} + \sqrt[3]{x+2} = 0$
$f(x) = \sqrt[3]{x-2} + \sqrt[3]{x} + \sqrt[3]{x+2}$
Firstly I check the amount of solutions.


*

*Graph of the function starts at the bottom and ends at the top.

*The derivative is always greater than 0, so the function is always growing.

*After all of that we know that there is just 1 solution.


Now I try to get the solution:


*

*$\sqrt[3]{x-2} + \sqrt[3]{x+2} = -\sqrt[3]{x} $

*$(\sqrt[3]{x-2} + \sqrt[3]{x+2})^3 = -x $

*$x-2 + x + 2 + 3\sqrt[3]{(x-2)^2(x+2)} + 3\sqrt[3]{(x-2)(x+2)^2} = -x$

*$3x + 3\sqrt[3]{(x-2)(x+2)}(\sqrt[3]{x-2} + \sqrt[3]{x+2})= 0$

*$3x + 3\sqrt[3]{(x-2)(x+2)}(-\sqrt[3]{x})= 0$

*$3(x - \sqrt[3]{x^2-4} * \sqrt[3]{x}) = 0$

*$\sqrt[3]{x^2} * \sqrt[3]{x} - \sqrt[3]{x^2-4} * \sqrt[3]{x} = 0$

*$\sqrt[3]{x}(\sqrt[3]{x^2} - \sqrt[3]{x^2-4}) = 0 $


So the solution is $x = 0$.
I am wondering if there is a faster way to do this, without checking the amount of solutions.

Observe that for the given function, $f(a)=-f(-a)$ for any value of $a$.
Now put $a=0$.
Thus $f(0)=-f(0)$.
=> $2f(0)=0$
=> $f(0)=0$
=> $x=0$ is a solution.
Also, $f'(x)>0$ for all real numbers $x$.
So, the function is always increasing and thus $x=0$ is the only solution.
