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Question Number 120748 by help last updated on 02/Nov/20

	Answered by TANMAY PANACEA last updated on 02/Nov/20

	$x = 3 y = - 2 z = - 1$
	Commented by help last updated on 02/Nov/20

	$s o l u t i o n p l e a s e$
	Answered by mindispower last updated on 02/Nov/20
	$ke_k =Σ_(i=1) ^k (−1)^(i−1) e_(k−i) p_i Newtoon identities e_k =Σ_(1≤a_1 .....a_k ≤n) X_a_1 .....X_a_k ...p_k =Σ_(i=1) ^n X_i ^k 2e_2 =e_1 p_1 −e_0 p_2 ⇒2e_2 =−p_2 3e_3 =e_2 p_1 −e_1 p_2 +p_3 ⇒e_3 =6 0=e_3 p_1 −e_2 p_2 +e_3 p_1 −p_4 =2e_2 ^2 −p_4 p_4 =2e_2 ^2 0=e_4 p_1 −e_3 p_2 +e_2 p_3 −e_1 p_4 +p_5 =−6p_2 +18e_2 +p_5 ⇒p_5 =−30e_2 0=e_5 p_1 −e_4 p_2 +e_3 p_3 −e_2 p_4 +e_1 p_5 −p_6 p_6 =108−2e_2 ^3 7e_7 =e_6 p_1 −e_5 p_2 +e_4 p_3 −e_3 p_4 +e_2 p_5 −e_1 p_6 +p_7 0=−6p_4 +e_2 (−30e_2 )+p_7 −42e_2 ^2 =−2058⇒e_2 =+_− (√((2058)/(42))) =+_− (√(49))=+_− 7 e_2 =7,e_1 =0,e_3 =6 X^3 +7X−6 cardan e_2 =−7 ⇒X^3 −7X−6,X=−1solution (X+1)(X^2 −X−6)⇒X∈{−1,3,−2} (x,y,z)∈{(−1,3,−2)} withe permutation$
	${k e}_{k} = \underset{i = 1}{\sum^{k}} {(- 1)}^{i - 1} e_{k - i} p_{i}$ $N e w t o o n i d e n t i t i e s$ $e_{k} = \sum_{1 ⩽ a_{1} . . . . . a_{k} ⩽ n} X_{a_{1}} . . . . . X_{a_{k}} . . . p_{k} = \underset{i = 1}{\sum^{n}} X_{i}^{k}$ $2 e_{2} = e_{1} p_{1} - e_{0} p_{2} \Rightarrow 2 e_{2} = - p_{2}$ $3 e_{3} = e_{2} p_{1} - e_{1} p_{2} + p_{3} \Rightarrow e_{3} = 6$ $0 = e_{3} p_{1} - e_{2} p_{2} + e_{3} p_{1} - p_{4} = 2 e_{2}^{2} - p_{4}$ $p_{4} = 2 e_{2}^{2}$ $0 = e_{4} p_{1} - e_{3} p_{2} + e_{2} p_{3} - e_{1} p_{4} + p_{5}$ $= - 6 p_{2} + 18 e_{2} + p_{5} \Rightarrow p_{5} = - 30 e_{2}$ $0 = e_{5} p_{1} - e_{4} p_{2} + e_{3} p_{3} - e_{2} p_{4} + e_{1} p_{5} - p_{6}$ $p_{6} = 108 - 2 e_{2}^{3}$ $7 e_{7} = e_{6} p_{1} - e_{5} p_{2} + e_{4} p_{3} - e_{3} p_{4} + e_{2} p_{5} - e_{1} p_{6} + p_{7}$ $0 = - 6 p_{4} + e_{2} (- 30 e_{2}) + p_{7}$ $- 42 e_{2}^{2} = - 2058 \Rightarrow e_{2} = \underline{+} \sqrt{\frac{2058}{42}}$ $= \underline{+} \sqrt{49} = \underline{+} 7$ $e_{2} = 7, e_{1} = 0, e_{3} = 6$ $X^{3} + 7 X - 6$ $c a r d a n$ $e_{2} = - 7$ $\Rightarrow X^{3} - 7 X - 6, X = - 1 s o l u t i o n$ $(X + 1) (X^{2} - X - 6) \Rightarrow X \in {- 1, 3, - 2}$ $(x, y, z) \in {(- 1, 3, - 2)} w i t h e p e r m u t a t i o n$
	Commented by help last updated on 02/Nov/20

	$G a v e k u d o s$
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