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Question Number 98293 by I want to learn more last updated on 12/Jun/20

	Answered by mr W last updated on 12/Jun/20

	$s e e Q 97675$ $x + y + z = 0$ $3 x y z = 18 \Rightarrow x y z = 6$ $7 {(x y + y z + z x)}^{2} (x y z) = 2058$ $\Rightarrow x y + y z + z x = \pm 7$ $x, y, z a r e r o o t s o f$ $t^{3} - 7 t - 6 = 0$ $(t + 1) (t + 2) (t - 3) = 0$ $t = - 1, - 2, 3$ $\Rightarrow x, y, z = - 1, - 2, 3$ $o r$ $t^{3} + 7 t - 6 = 0$ $t_{1} = \sqrt[3]{3 + \sqrt{9 + {(\frac{7}{3})}^{3}}} + \sqrt[3]{3 - \sqrt{9 + {(\frac{7}{3})}^{3}}}$ $t_{2, 3} = c o m p l e x$
	Commented by I want to learn more last updated on 12/Jun/20

	$wow, thanks sir .$
	Answered by 1549442205 last updated on 13/Jun/20
	$we have z=−(x+y).Replace into ii)we get x^3 +y^3 −(x+y)^3 =18⇔x^3 +y^3 −(x^3 +3xy(x+y)+y^3 )=18 ⇔−3xy(x+y)=18⇔3xyz=18⇔xyz=6 Replace into iii)we get x^7 +y^7 −(x+y)^7 =2058(∗) we have:x^3 +y^3 =(x+y)^3 −3xy(x+y)=−z^3 +3xyz=18−z^3 (1) x^5 +y^5 =(x+y)^5 −5xy(x^3 +y^3 )−10x^2 y^2 (x+y) =−z^5 +5xyz^3 −90xy+10x^2 y^2 z=−z^5 +30z^2 −30xy(2) (x+y)^7 =x^7 +y^7 +7xy(x^5 +y^5 )+21x^2 y^2 (x^3 +y^3 )+35x^3 y^3 (x+y) =x^7 +y^7 −42z^4 +1260z−210x^2 y^2 −756z+378x^2 y^2 −210x^2 y^2 (3). Replace into (∗)we get : 42z^4 +42x^2 y^2 −504z=2058⇔z^4 +x^2 y^2 −12z=49 z^4 +((36)/z^2 )−12z−49=0⇔z^6 −12z^3 −49z^2 +36=0 ⇔(z+1)(z+2)(z−3)(z^3 +7z−6)=0 ⇔z∈{−1;−2;3;((^3 (√(81+3(√(1758)))))/3)+((−7^3 (√2))/(^3 (√(162+6(√(1758))))))} a/for z∈{−1;−2;3} We get (x;y;z)={(−1;−2;3);(−1;3;−2);(−2;−1;3);(−2;3;−1) (3;−1;−2);(3;−2;−1)} b/for z=((^3 (√(81+3(√(1758)))))/3)+((−7^3 (√2))/(^3 (√(162+6(√(1758)))))) x,y are image roots$
	$we have z = - (x + y) . Replace into ii) we get$ $x^{3} + y^{3} - {(x + y)}^{3} = 18 \Leftrightarrow x^{3} + y^{3} - (x^{3} + 3 xy (x + y) + y^{3}) = 18$ $\Leftrightarrow - 3 xy (x + y) = 18 \Leftrightarrow 3 xyz = 18 \Leftrightarrow xyz = 6$ $Replace into iii) we get x^{7} + y^{7} - {(x + y)}^{7} = 2058 ()$ $we have : x^{3} + y^{3} = {(x + y)}^{3} - 3 xy (x + y) = - z^{3} + 3 xyz = 18 - z^{3} (1)$ $x^{5} + y^{5} = {(x + y)}^{5} - 5 xy (x^{3} + y^{3}) - 10 x^{2} y^{2} (x + y)$ $= - z^{5} + 5 {xyz}^{3} - 90 xy + 10 x^{2} y^{2} z = - z^{5} + 30 z^{2} - 30 xy (2)$ ${(x + y)}^{7} = x^{7} + y^{7} + 7 xy (x^{5} + y^{5}) + 21 x^{2} y^{2} (x^{3} + y^{3}) + 35 x^{3} y^{3} (x + y)$ $= x^{7} + y^{7} - 42 z^{4} + 1260 z - 210 x^{2} y^{2} - 756 z + 378 x^{2} y^{2} - {210 x}^{2} y^{2} (3) .$ $Replace into () we get :$ $42 z^{4} + 42 x^{2} y^{2} - 504 z = 2058 \Leftrightarrow z^{4} + x^{2} y^{2} - 12 z = 49$ $z^{4} + \frac{36}{z^{2}} - 12 z - 49 = 0 \Leftrightarrow z^{6} - 12 z^{3} - 49 z^{2} + 36 = 0$ $\Leftrightarrow (z + 1) (z + 2) (z - 3) (z^{3} + 7 z - 6) = 0$ $\Leftrightarrow z \in {- 1; - 2; 3; \frac{^{3} \sqrt{81 + 3 \sqrt{1758}}}{3} + \frac{- 7^{3} \sqrt{2}}{^{3} \sqrt{162 + 6 \sqrt{1758}}}}$ $a / for z \in {- 1; - 2; 3}$ $We get (x; y; z) = {(- 1; - 2; 3); (- 1; 3; - 2); (- 2; - 1; 3); (- 2; 3; - 1)$ $(3; - 1; - 2); (3; - 2; - 1)}$ $b / for z = \frac{^{3} \sqrt{81 + 3 \sqrt{1758}}}{3} + \frac{- 7^{3} \sqrt{2}}{^{3} \sqrt{162 + 6 \sqrt{1758}}}$ $x, y are image roots$
	Commented by I want to learn more last updated on 13/Jun/20

	$Wow, thanks sir$
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