Why four roots to this equation: $(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$
Consider this system of polynomial equations:
$$\begin{align}
a+b+c &= 2 \\
a^3 &= 7x + 1 \\
b^3 &= 8 + x - x^2 \\
c^3 &= x^2 - 8 x - 1
\end{align}$$
Using the method of resultants (via the oh-so-convenient Resultant[] function in Mathematica), we can eliminate $a$, $b$, $c$ to get
$$x^3\;(x-1)^6\;(x+1)^3\;(x - 9 )^3 \;=\; 0$$
Therefore, the only candidate values for $x$ are, as you have previously determined, $0$, $1$, $-1$, and $9$. Sometimes the method of resultants introduces extraneous factors, so we should double-check. As it turns out (assuming the $1/3$ power indicates the real cube root of a real number), each of the four candidates is in fact a valid solution. $\square$
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.
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.