Why four roots to this equation: $(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$ 
$$(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$$

I figured the roots are $0$, $1$, $-1$, and $9$. But why?

Let: $(7x+1)^{1/3} = a,(x^2-8x-1)^{1/3} = b, (8+x-x^2)^{1/3} = c.$
You have: $$\begin{cases} a+b+c = 2 \\ a^3+b^3+c^3 = 8 \end{cases} \Rightarrow a^3+b^3+c^3 = (a+b+c)^3$$
$$ \Leftrightarrow a^3+b^3 = (a+b+c)^3-c^3 = (a+b)[(a+b+c)^2+c(a+b+c)+c^2)]$$
$$ \Leftrightarrow a+b = 0 \text{ or }a^2-ab+b^2 = (a+b+c)^2+c(a+b+c)+c^2 $$
The first case is easy: $a = -b \Leftrightarrow a^3 = -b^3 \Leftrightarrow 7x+1 = x^2-8x-1 \Leftrightarrow ... $
The second case is equivalent to:
$$ab+c(a+b)+c^2 = 0 \Leftrightarrow (a+c)(b+c) = 0$$
I guess you can finish the rest.
