\(\frac {(x- 8)^4 (x - 2)^5 (x - 1)^3 (x + 1)^2 (x + 5)^2}{x^4 - 2x^3 - x + 2} \le 0\)
⇒ \(\frac {(x- 8)^4 (x - 2)^5 (x - 1)^3 (x + 1)^2 (x + 5)^2}{(x - 1)(x - 2)(x^2 + x + 1)} \le 0\)
As (x - 1) (x - 2) (x2 + x + 1) are in denominator
So,
\(x^2 + x + 1 > 0\, \forall x\)
\(x - 1 \ne 0 \) ⇒ \(x \ne 1\)
& \(x - 2 \ne 0 \) ⇒ \(x \ne 2\)
⇒ \(\frac {(x- 8)^4 (x - 2)^4 (x - 1)^2 (x + 1)^2 (x + 5)^2}{x^2 + x + 1} \le 0, x\ne 1,2\)
\(\because (x - 8)^4 \ge 0 , (x - 2)^4 \ge 0 , (x- 1)^2 \ge 0\)
\((x + 1)^2\ge 0 , (x + 5)^2 \ge 0 \) & \(x^2 + x + 1 > 0\)
\(\therefore\) In equality holds only if
\((x - 8)^2 (x - 2)^4 (x - 1)^2 (x + 1)^2 (x + 5)^2 = 0, x \ne 1,2\)
⇒ \((x - 8) (x + 1) (x + 5) = 0\) (As x \(\ne\) 1 & x \(\ne\) 2)
⇒ \(x = 8 , - 1 , -5\)
Sum of all integral solutions \(= 8 + (-1) + (-5)\)
\(= 8 - 6\)
\(= 2\)