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Question Number 187811 by Rupesh123 last updated on 22/Feb/23

	Answered by Frix last updated on 22/Feb/23
	$x^6 +(x^5 /2)−((5x^4 )/4)−((5x^3 )/8)+((5x^2 )/(16))+((5x)/(32))+(1/(64))=0 Try to find 2 cubic factors. If it′s possible to solve I guess the constant factors must be ±(√(1/(64)))=±(1/8): let s∈{−1, 1} (x^3 +ax^2 +bx+(s/8))(x^3 +cx^2 +dx+(s/8))=0 By matching the coefficients we get a=0 b=−(3/4) c=(1/2) d=−(1/2) s=−1 ⇒ (x^3 −((3x)/4)−(1/8))(x^3 +(x^2 /2)−(x/2)−(1/8))=0 We can solve these: x_1 =−cos ((2π)/9) x_2 =−sin (π/(18)) x_3 =cos (π/9) x_4 =−cos (π/7) x_5 =−sin (π/(14)) x_6 =sin ((3π)/(14))$
	$x^{6} + \frac{x^{5}}{2} - \frac{5 x^{4}}{4} - \frac{5 x^{3}}{8} + \frac{5 x^{2}}{16} + \frac{5 x}{32} + \frac{1}{64} = 0$ $Try to find 2 cubic factors . If {it}^{'} s possible$ $to solve I guess the constant factors must$ $be \pm \sqrt{\frac{1}{64}} = \pm \frac{1}{8} : let s \in {- 1, 1}$ $(x^{3} + {a x}^{2} + b x + \frac{s}{8}) (x^{3} + {c x}^{2} + d x + \frac{s}{8}) = 0$ $By matching the coefficients we get$ $a = 0 b = - \frac{3}{4} c = \frac{1}{2} d = - \frac{1}{2} s = - 1$ $\Rightarrow$ $(x^{3} - \frac{3 x}{4} - \frac{1}{8}) (x^{3} + \frac{x^{2}}{2} - \frac{x}{2} - \frac{1}{8}) = 0$ $We can solve these :$ $x_{1} = - \cos \frac{2 π}{9} x_{2} = - \sin \frac{π}{18} x_{3} = \cos \frac{π}{9}$ $x_{4} = - \cos \frac{π}{7} x_{5} = - \sin \frac{π}{14} x_{6} = \sin \frac{3 π}{14}$
	Commented by Rupesh123 last updated on 22/Feb/23
	Great!
	Commented by Frix last updated on 22/Feb/23
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