Let x = cos(π/9) Show that 8x^3 −6x−1=0 Deduce x is not rational


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Question Number 207194 by sniper237 last updated on 09/May/24

$L e t x = c o s \frac{π}{9}$ $S h o w t h a t 8 x^{3} - 6 x - 1 = 0$ $D e d u c e x i s n o t r a t i o n a l$
	Answered by Berbere last updated on 09/May/24
	$cos(3.(π/9))=cos((π/3))=(1/2) cos(3a)=Re(e^(ia) )^3 =4cos^3 (a)−3cos(a) ⇒(1/2)=4x^3 −3x⇒8x^3 −6x−1=0 suupose x=(p/q);gcd(p,q)=1; ⇒8p^3 −6pq^2 −q^3 =0⇔p(8p^2 −6q^2 )=q^3 ⇒p∣q^3 ⇒(gcd(p,q)=1⇒gcd(p,q^3 )=1⇒p=1 ⇒8−6q^2 −q^3 =0⇒q^2 ∣8⇒q∈{1,2} x=1;x=(1/2) impossible ⇒x∉IQ$
	$c o s (3 . \frac{π}{9}) = c o s (\frac{π}{3}) = \frac{1}{2}$ $c o s (3 a) = R e {(e^{i a})}^{3} = 4 {c o s}^{3} (a) - 3 c o s (a)$ $\Rightarrow \frac{1}{2} = 4 x^{3} - 3 x \Rightarrow 8 x^{3} - 6 x - 1 = 0$ $s u u p o s e x = \frac{p}{q}; g c d (p, q) = 1;$ $\Rightarrow 8 p^{3} - 6 {p q}^{2} - q^{3} = 0 \Leftrightarrow p (8 p^{2} - 6 q^{2}) = q^{3}$ $\Rightarrow p ∣ q^{3} \Rightarrow (g c d (p, q) = 1 \Rightarrow g c d (p, q^{3}) = 1 \Rightarrow p = 1$ $\Rightarrow 8 - 6 q^{2} - q^{3} = 0 \Rightarrow q^{2} ∣ 8 \Rightarrow q \in {1, 2} x = 1; x = \frac{1}{2} i m p o s s i b l e$ $\Rightarrow x \notin IQ$
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