<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] 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. $\square$
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