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(D) The given equation is $x^5-5x^4+9x^3-9x^2+5x-1=0$. We can rewrite this as $(x^5-1) - 5x(x^3-1) + 9x^2(x-1) = 0$. Factoring out $(x-1)$,we get $(x-

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$\sqrt{5}$

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(D) The given equation is $x^5-5x^4+9x^3-9x^2+5x-1=0$. We can rewrite this as $(x^5-1) - 5x(x^3-1) + 9x^2(x-1) = 0$. Factoring out $(x-1)$,we get $(x-1)(x^4+x^3+x^2+x+1) - 5x(x-1)(x^2+x+1) + 9x^2(x-1) = 0$. Dividing by $(x-1)$ (assuming $x 
eq 1$),we get $x^4+x^3+x^2+x+1 - 5x^3-5x^2-5x + 9x^2 = 0$. Simplifying,$x^4-4x^3+5x^2-4x+1 = 0$. Dividing by $x^2$,we get $x^2-4x+5-4/x+1/x^2 = 0$. Grouping terms,$(x^2+1/x^2) - 4(x+1/x) + 5 = 0$. Let $t = x+1/x$,then $t^2-2-4t+5 = 0$,so $t^2-4t+3 = 0$. Solving for $t$,$(t-3)(t-1) = 0$,so $t=3$ or $t=1$. For $t=3$,$x+1/x=3 \implies x^2-3x+1=0$,roots are $x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$. The difference of these roots is $\frac{3+\sqrt{5}}{2} - \frac{3-\sqrt{5}}{2} = \sqrt{5}$. For $t=1$,$x+1/x=1 \implies x^2-x+1=0$,which has complex roots.

The difference of the irrational roots of the equation $x^5-5x^4+9x^3-9x^2+5x-1=0$ is

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