One of the roots of the equation x^{14}+x^9-x | vedclass

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(D) Given equation: $x^{14}+x^9-x^5-1=0$ Factorizing the expression: $x^9(x^5+1) - 1(x^5+1) = 0$ $(x^9-1)(x^5+1) = 0$ This implies $x^5 = -1$ or $x^9

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$\frac{\sqrt{5}+1}{4}+i\frac{\sqrt{10-2\sqrt{5}}}{4}$

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(D) Given equation: $x^{14}+x^9-x^5-1=0$ Factorizing the expression: $x^9(x^5+1) - 1(x^5+1) = 0$ $(x^9-1)(x^5+1) = 0$ This implies $x^5 = -1$ or $x^9 = 1$. For $x^5 = -1$,we have $x^5 = \cos(180^{\circ}) + i\sin(180^{\circ})$. The roots are given by $x = \cos(\frac{180^{\circ}+360^{\circ}k}{5}) + i\sin(\frac{180^{\circ}+360^{\circ}k}{5})$ for $k=0, 1, 2, 3, 4$. For $k=0$,$x = \cos(36^{\circ}) + i\sin(36^{\circ})$. Using trigonometric values,$\cos(36^{\circ}) = \frac{\sqrt{5}+1}{4}$ and $\sin(36^{\circ}) = \frac{\sqrt{10-2\sqrt{5}}}{4}$. Thus,$x = \frac{\sqrt{5}+1}{4} + i\frac{\sqrt{10-2\sqrt{5}}}{4}$ is a root.

One of the roots of the equation $x^{14}+x^9-x^5-1=0$ is

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