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Let-x-cos-pi-9-Show-that-8x-3-6x-1-0-Deduce-x-is-not-rational-




Question Number 207194 by sniper237 last updated on 09/May/24
Let  x = cos(π/9)   Show that 8x^3 −6x−1=0  Deduce x is not  rational
$${Let}\:\:{x}\:=\:{cos}\frac{\pi}{\mathrm{9}}\: \\ $$$${Show}\:{that}\:\mathrm{8}{x}^{\mathrm{3}} −\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$${Deduce}\:{x}\:{is}\:{not}\:\:{rational}\: \\ $$
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
$${cos}\left(\mathrm{3}.\frac{\pi}{\mathrm{9}}\right)={cos}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\mathrm{3}{a}\right)={Re}\left({e}^{{ia}} \right)^{\mathrm{3}} =\mathrm{4}{cos}^{\mathrm{3}} \left({a}\right)−\mathrm{3}{cos}\left({a}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}\Rightarrow\mathrm{8}{x}^{\mathrm{3}} −\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$${suupose}\:{x}=\frac{{p}}{{q}};{gcd}\left({p},{q}\right)=\mathrm{1}; \\ $$$$\Rightarrow\mathrm{8}{p}^{\mathrm{3}} −\mathrm{6}{pq}^{\mathrm{2}} −{q}^{\mathrm{3}} =\mathrm{0}\Leftrightarrow{p}\left(\mathrm{8}{p}^{\mathrm{2}} −\mathrm{6}{q}^{\mathrm{2}} \right)={q}^{\mathrm{3}} \\ $$$$\Rightarrow{p}\mid{q}^{\mathrm{3}} \Rightarrow\left({gcd}\left({p},{q}\right)=\mathrm{1}\Rightarrow{gcd}\left({p},{q}^{\mathrm{3}} \right)=\mathrm{1}\Rightarrow{p}=\mathrm{1}\right. \\ $$$$\Rightarrow\mathrm{8}−\mathrm{6}{q}^{\mathrm{2}} −{q}^{\mathrm{3}} =\mathrm{0}\Rightarrow{q}^{\mathrm{2}} \mid\mathrm{8}\Rightarrow{q}\in\left\{\mathrm{1},\mathrm{2}\right\}\:{x}=\mathrm{1};{x}=\frac{\mathrm{1}}{\mathrm{2}}\:{impossible} \\ $$$$\Rightarrow{x}\notin\mathbb{IQ} \\ $$

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