If alpha and beta are two complex roots of th | vedclass

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(D) Given equation: $6x^6-25x^5+31x^4-31x^2+25x-6=0$ Rearranging terms: $6(x^6-1) - 25x(x^4-1) + 31x^2(x^2-1) = 0$ Factoring $(x^2-1)$: $(x^2-1)[6(x^4

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$\frac{5}{3}$

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(D) Given equation: $6x^6-25x^5+31x^4-31x^2+25x-6=0$ Rearranging terms: $6(x^6-1) - 25x(x^4-1) + 31x^2(x^2-1) = 0$ Factoring $(x^2-1)$: $(x^2-1)[6(x^4+x^2+1) - 25x(x^2+1) + 31x^2] = 0$ $(x^2-1)[6x^4 - 25x^3 + 37x^2 - 25x + 6] = 0$ Dividing the second bracket by $x^2$: $x^2(x^2-1)[6(x^2+\frac{1}{x^2}) - 25(x+\frac{1}{x}) + 37] = 0$ Let $y = x+\frac{1}{x}$,then $x^2+\frac{1}{x^2} = y^2-2$. $6(y^2-2) - 25y + 37 = 0 \Rightarrow 6y^2 - 25y + 25 = 0$ $(2y-5)(3y-5) = 0 \Rightarrow y = \frac{5}{2}, \frac{5}{3}$ For $x+\frac{1}{x} = \frac{5}{2}$,$x=2, \frac{1}{2}$ (real roots). For $x+\frac{1}{x} = \frac{5}{3}$,$3x^2-5x+3=0$. The roots are complex. Sum of these complex roots $\alpha+\beta = -\frac{b}{a} = -(\frac{-5}{3}) = \frac{5}{3}$.

If $\alpha$ and $\beta$ are two complex roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$,then $\alpha+\beta=$

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