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Question Number 188645 by Rupesh123 last updated on 04/Mar/23

	Answered by witcher3 last updated on 05/Mar/23
	$let a=cos((π/7)),b=−cos(((2π)/7));c=cos(((3π)/7)) we want 1+a^3 +b^3 +c^3 +a^3 b^3 +a^3 c^3 +b^3 c^3 +a^3 b^3 c^3 =S Starte withe eq∵cos(3x)=−cos(4x)⇒cos(3x)=cos(π−4x) ⇔3x=π−4x+2kπ∨3x=4x−π+2kπ ⇔x_k =(π/7)(1+2k),k∈[0,6] cos(3x)=4cos^3 (x)−3cos(x) cos(4x)=2(2cos^2 (x)−1)^2 −1=8cos^4 (x)−8cos^2 (x)+1 ⇔−8cos^4 (x)−4cos^3 (x)+8cos^2 (x)+3cos(x)−1=0 cos(x_k ) root of cos((π/7)),cos(((3π)/7)),cos(((5π)/7))=−cos(((2π)/7)),cosπ=−1 cos(((9π)/7))=−cos(((2π)/7)),cos((11π)/7)=cos(((3π)/7));cos(((13π)/7))=cos((π/7)) ⇔{−1,cos(π/7),−cos(((2π)/7)),cos(((3π)/7))}root of −8X^4 −4X^3 +8x^2 +3x−1=0..(−1)root ⇔8x^4 −4x^3 +8x^2 +3x−1=(x+1)(−8x^3 +4x^2 +4x−1) ⇔{cos(π/7),−cos(((2π)/7)),cos(((3π)/7))}root of x^3 −(x^2 /2)+(x/2)+(1/8) a+b+c=(1/2),ab+bc+ac=−(1/2),abc=−(1/8) p_3 =a^3 +b^3 +c^3 ..Newtoon identie p_2 =e_1 p_1 −2e_2 =(1/4)+1=(5/4) p_3 =p_2 e_1 −e_2 p_1 +3e_3 =(5/8)+(1/4)−(3/8)=(1/2) ab+ac+bc=−(1/2) ab.ac+acbc+ab.bc=abc(a+b+c)=−(1/(16)) (abc)^2 =(1/(64)) p_2 =(1/4)+(1/8)=(3/8) p_3 ′=(3/8).−(1/2)−(1/(16)).(1/2)+(3/(64))=−((11)/(64)) P_3 ′′=(abc)^3 =−(1/(512)) S=1+p_3 +p_3 ′+p′′_3 =1+(1/2)−((11)/(64))−(1/(512)) =((85)/(64))−(1/(512))=((679)/(512)) ⇔Π_(k=0) ^2 (1+cos^3 (((2k+1)/7)π))=((679)/(512))$
	$let a = \cos (\frac{π}{7}), b = - \cos (\frac{2 π}{7}); c = \cos (\frac{3 π}{7})$ $we want 1 + a^{3} + b^{3} + c^{3} + a^{3} b^{3} + a^{3} c^{3} + b^{3} c^{3} + a^{3} b^{3} c^{3} = S$ $Starte withe$ $eq ∵ \cos (3 x) = - \cos (4 x) \Rightarrow \cos (3 x) = \cos (π - 4 x)$ $\Leftrightarrow 3 x = π - 4 x + 2 k π \lor 3 x = 4 x - π + 2 k π$ $\Leftrightarrow x_{k} = \frac{π}{7} (1 + 2 k), k \in [0, 6]$ $\cos (3 x) = {4 \cos}^{3} (x) - 3 \cos (x)$ $\cos (4 x) = 2 {({2 \cos}^{2} (x) - 1)}^{2} - 1 = 8 \cos^{4} (x) - 8 \cos^{2} (x) + 1$ $\Leftrightarrow - {8 \cos}^{4} (x) - {4 \cos}^{3} (x) + {8 \cos}^{2} (x) + 3 \cos (x) - 1 = 0$ $\cos (x_{k}) root of$ $\cos (\frac{π}{7}), \cos (\frac{3 π}{7}), \cos (\frac{5 π}{7}) = - \cos (\frac{2 π}{7}), \cos π = - 1$ $\cos (\frac{9 π}{7}) = - \cos (\frac{2 π}{7}), \cos \frac{11 π}{7} = \cos (\frac{3 π}{7}); \cos (\frac{13 π}{7}) = \cos (\frac{π}{7})$ $\Leftrightarrow {- 1, \cos \frac{π}{7}, - \cos (\frac{2 π}{7}), \cos (\frac{3 π}{7})} root of$ $- {8 X}^{4} - {4 X}^{3} + {8 x}^{2} + 3 x - 1 = 0 . . (- 1) root$ $\Leftrightarrow {8 x}^{4} - {4 x}^{3} + {8 x}^{2} + 3 x - 1 = (x + 1) (- {8 x}^{3} + {4 x}^{2} + 4 x - 1)$ $\Leftrightarrow {\cos \frac{π}{7}, - \cos (\frac{2 π}{7}), \cos (\frac{3 π}{7})} root of$ $x^{3} - \frac{x^{2}}{2} + \frac{x}{2} + \frac{1}{8}$ $a + b + c = \frac{1}{2}, ab + bc + ac = - \frac{1}{2}, abc = - \frac{1}{8}$ $p_{3} = a^{3} + b^{3} + c^{3} . . Newtoon identie$ $p_{2} = e_{1} p_{1} - {2 e}_{2} = \frac{1}{4} + 1 = \frac{5}{4}$ $p_{3} = p_{2} e_{1} - e_{2} p_{1} + {3 e}_{3} = \frac{5}{8} + \frac{1}{4} - \frac{3}{8} = \frac{1}{2}$ $ab + ac + bc = - \frac{1}{2}$ $ab . ac + acbc + ab . bc = abc (a + b + c) = - \frac{1}{16}$ ${(abc)}^{2} = \frac{1}{64}$ $p_{2} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$ $p_{3}^{'} = \frac{3}{8} . - \frac{1}{2} - \frac{1}{16} . \frac{1}{2} + \frac{3}{64} = - \frac{11}{64}$ $P_{3}^{″} = {(abc)}^{3} = - \frac{1}{512}$ $S = 1 + p_{3} + p_{3}^{'} + p_{3}^{″} = 1 + \frac{1}{2} - \frac{11}{64} - \frac{1}{512}$ $= \frac{85}{64} - \frac{1}{512} = \frac{679}{512}$ $\Leftrightarrow \underset{k = 0}{\prod^{2}} (1 + \cos^{3} (\frac{2 k + 1}{7} π)) = \frac{679}{512}$
	Commented by Rupesh123 last updated on 06/Mar/23
	Good job, sir!
	Commented by witcher3 last updated on 06/Mar/23

	$thank You Sir$ $god bless You$
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