\begin{align} & (x-1)(x-2)(x-3) + (x-2)(x-3)(x-4) \\ + & (x-3)(x-4)(x-5) + (x-4)(x-5)(x-6) \\ + & (x-5)(x-6)(x-7) + (x-6)(x-7)(x-8) =0 ? \end{align}
To find the value(s) of x that satisfy the equation
\begin{align} & (x-1)(x-2)(x-3) + (x-2)(x-3)(x-4) \\ + & (x-3)(x-4)(x-5) + (x-4)(x-5)(x-6) \\ + & (x-5)(x-6)(x-7) + (x-6)(x-7)(x-8) =0, \\ \end{align}
we can first simplify the equation by expanding each term and then combining like terms. Let's go step by step:
Expanding the first term:
\begin{align}
(x-1)(x-2)(x-3) &= (x^2 - 3x + 2x - 2)(x-3) \\
&= (x^2 - x - 2)(x-3) \\
&= x^3 - 3x^2 - x^2 + 3x - 2x + 6 \\
&= x^3 - 4x^2 + x + 6.
\end{align}
Expanding the second term:
\begin{align}
(x-2)(x-3)(x-4) &= (x^2 - 3x + 2x - 6)(x-4) \\
&= (x^2 - x - 6)(x-4) \\
&= x^3 - 4x^2 - 6x - 4x^2 + 16x + 24 \\
&= x^3 - 8x^2 + 10x + 24.
\end{align}
Expanding the third term:
\begin{align}
(x-3)(x-4)(x-5) &= (x^2 - 4x + 3x - 12)(x-5) \\
&= (x^2 - x - 12)(x-5) \\
&= x^3 - 5x^2 - 12x - x^2 + 5x + 60 \\
&= x^3 - 6x^2 - 7x + 60.
\end{align}
Expanding the fourth term:
\begin{align}
(x-4)(x-5)(x-6) &= (x^2 - 5x + 4x - 20)(x-6) \\
&= (x^2 - x - 20)(x-6) \\
&= x^3 - 6x^2 - 20x - x^2 + 6x + 120 \\
&= x^3 - 7x^2 - 14x + 120.
\end{align}
Expanding the fifth term:
\begin{align}
(x-5)(x-6)(x-7) &= (x^2 - 6x + 5x - 30)(x-7) \\
&= (x^2 - x - 30)(x-7) \\
&= x^3 - 7x^2 - 30x - x^2 + 7x + 210 \\
&= x^3 - 8x^2 - 23x + 210.
\end{align}
Expanding the sixth term:
\begin{align}
(x-6)(x-7)(x-8) &= (x^2 - 7x + 6x - 42)(x-8) \\
&= (x^2 - x - 42)(x-8) \\
&= x^3 - 8x^2 - 42x - x^2 + 8x + 336 \\
&= x^3 - 9x^2 - 34x + 336.
\end{align}
Now we can substitute all the expanded terms back into the original equation and combine like terms:
\begin{align}
x^3 - 4x^2 + x + 6 + x^3 - 8x^2 + 10x + 24 + x^3 - 6x^2 - 7x + 60 + \\ x^3 - 7x^2 - 14x + 120 + x^3 - 8x^2 - 23x + 210 + x^3 - 9x^2 - 34x + 336 &= 0.
\end{align}
Combining like terms:
\begin{align}
6x^3 - 42x^2 - 47x + 756 &= 0.
\end{align}
Unfortunately, this equation is not easily solvable by factoring or simple manipulation. In this case, we can use numerical methods such as graphing or iterative methods (like Newton's method) to approximate the solutions.