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(C) The given equation is $6x^6-25x^5+31x^4-31x^2+25x-6=0$. This is a reciprocal equation of the first type. Dividing by $x^3$,we get $6(x^3 - \frac{1

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$3$

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(C) The given equation is $6x^6-25x^5+31x^4-31x^2+25x-6=0$. This is a reciprocal equation of the first type. Dividing by $x^3$,we get $6(x^3 - \frac{1}{x^3}) - 25(x^2 + \frac{1}{x^2}) + 31(x - \frac{1}{x}) = 0$. Let $t = x - \frac{1}{x}$. Then $x^2 + \frac{1}{x^2} = t^2 + 2$ and $x^3 - \frac{1}{x^3} = t^3 + 3t$. Substituting these into the equation: $6(t^3 + 3t) - 25(t^2 + 2) + 31t = 0$. $6t^3 - 25t^2 + 49t - 50 = 0$. By testing rational roots,$t=2$ is a root: $6(8) - 25(4) + 49(2) - 50 = 48 - 100 + 98 - 50 = -4 
eq 0$. Actually,checking $t=2$: $6(8) - 25(4) + 49(2) - 50 = 48 - 100 + 98 - 50 = -4$. Let's re-examine the equation: $6(x^6-1) - 25x(x^4-1) + 31x^2(x^2-1) = 0$. $(x^2-1)[6(x^4+x^2+1) - 25x(x^2+1) + 31x^2] = 0$. $(x-1)(x+1)(6x^4 - 25x^3 + 37x^2 - 25x + 6) = 0$. The roots $x=1$ and $x=-1$ are rational. For $6x^4 - 25x^3 + 37x^2 - 25x + 6 = 0$,divide by $x^2$: $6(x^2 + \frac{1}{x^2}) - 25(x + \frac{1}{x}) + 37 = 0$. Let $u = x + \frac{1}{x}$. Then $6(u^2-2) - 25u + 37 = 0 \implies 6u^2 - 25u + 25 = 0$. $(2u-5)(3u-5) = 0$,so $u = \frac{5}{2}$ or $u = \frac{5}{3}$. If $x + \frac{1}{x} = \frac{5}{2}$,$2x^2 - 5x + 2 = 0 \implies (2x-1)(x-2) = 0$,so $x = 2, \frac{1}{2}$. If $x + \frac{1}{x} = \frac{5}{3}$,$3x^2 - 5x + 3 = 0$,which has no real roots. The rational roots are $\{-1, 1, \frac{1}{2}, 2\}$. The smallest is $m = -1$ and the greatest is $M = 2$. Thus,$M-m = 2 - (-1) = 3$.

If $m$ and $M$ are respectively the smallest and greatest rational roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$,then $M-m=$

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