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\(\frac {(x- 8)^4 (x - 2)^5 (x - 1)^3 (x + 1)^2 (x + 5)^2}{x^4 - 2x^3 - x + 2} \le 0\)

⇒ \(\frac {(x- 8)^4 (x - 2)^5 (x - 1)^3 (x + 1)^2 (x + 5)^2}{(x - 1)(x - 2)(x^2 + x + 1)} \le 0\)

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 \(= 8 + (-1) + (-5)\) 

\(=  8 - 6\)

\(= 2\)

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