<s>[INST] 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?
 [/INST] Why does this equation has $4$ roots? The short answer is it is an accident.
For similar problem like
$$(7x-6)^{1/3}+(8+x-x^2)^{1/3}+(x^2-8x-1)^{1/3}=1
$$
You can get $5$ instead of $4$ solutions.
Define $a,b,c$ like Blue's answer.
$$\begin{cases}
a^3 &= 7x + 1 \\
b^3 &= 8 + x - x^2 \\
c^3 &= x^2 - 8 x - 1
\end{cases}
$$
The key of the whole thing is $a^3 + b^3 + c^3 = 8$ which is independent of $x$
and $8 = 2^3$.  
If $a + b + c = 2$, we will have
$$0 = (a+b+c)^3 - (a^3+b^3+c^3) = 3(a+b)(b+c)(c+a) = 3(2-c)(2-a)(2-b)$$
This implies one of $a, b, c$ is equal to $2$ and hence one of $a^3, b^3, c^3$ is $8$.


*

*$a^3 = 8 \implies x = 1$,

*$b^3 = 8 \implies x - x^2 = 0 \iff x(x-1) = 0 \implies x = 0 \text{ or } 1$.

*$c^3 = 8 \implies x^2 - 8x = 9  \iff (x-9)(x+1) = 0 \implies x = -1\text{ or } 9$.


As you can see, the first case is a linear equation, it give you one possible solution.
The remaining two cases are quadratic equation, each of them given you two possible solutions. This means in principle, you can have $5$ possible solutions.
However, one of the solution $x = 1$ has been duplicated, so you are left with $4$ distinct possible solutions.
It remains to check the above 4 possible choice of $x$ are indeed solutions of the
original problem. Instead of plugging in the values of $x$ to check it is a solution, let me 
use $x = 0$ as an example to illustrate the general pattern. In that case $b^3 = 8$ and we have
$$a^3 + b^3 + c^3 = 8 \implies a^3 + c^3 = (a+c)(a^2 - ac + c^2) = 0$$
Notice $a, c$ are real numbers, this means
$a^2 - ac + c^2 = (a-\frac{c}{2})^2 + \frac34 c^2 \ne 0$ whenever $(a,c) \ne 0$. 
This implies $ a+c = 0$ and hence
$$(a + b + c)^3 = a^3 + b^3 + c^3 = 8 \implies a + b + c = 2$$
As a result, $x = 0$ is indeed a solution. By a similar argument, the other $3$ choices of $x$ are also solutions of the problem.
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