Advanced Equation #1

Let $a=\sqrt[3]{x+45}$ and $b=\sqrt[3]{16-x}$ .

Then you'll see that

$a^3+b^3=61\quad \text{and} \quad a+b=1$

Since $(a+b)^3=a^3+b^3+3ab(a+b)$ ,

$a+b=1 \quad \text{and} \quad ab=-20$

Therefore $a=5$ or $a=-4$ .

Substituting to the definition of $a$ , we get:

$x=80$ or $x=-109$ .

Since $\sqrt[3]{125}+\sqrt[3]{-64}=1$ ,

The sum of all real roots of the given equation is $\boxed{-29}$ .