{ ((x^2 +y^2 +z^2 =70)),((x^3 +y^3 +z^3 =64)),((x^4 +y^4 +z^4 =2002)),(((x+y)(y+z)(z+x)=?)) :} (Use Newton-Identities or otherwise)


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Question Number 166009 by Rasheed.Sindhi last updated on 11/Feb/22
${ ((x^2 +y^2 +z^2 =70)),((x^3 +y^3 +z^3 =64)),((x^4 +y^4 +z^4 =2002)),(((x+y)(y+z)(z+x)=?)) :} (Use Newton-Identities or otherwise)$
${\begin{cases} x^{2} + y^{2} + z^{2} = 70 \\ x^{3} + y^{3} + z^{3} = 64 \\ x^{4} + y^{4} + z^{4} = 2002 \\ (x + y) (y + z) (z + x) = ? \end{cases}$ $(U s e N e w t o n - I d e n t i t i e s$ $o r o t h e r w i s e)$
Commented by mr W last updated on 11/Feb/22

$i n r e v e r s e d d i r e c t i o n t h e r e a r e m u l t i p l e$ $s o l u t i o n s .$ $t h e e q u a t i o n s y s t e m h a s 24 s o l u t i o n$ $t r i p l e s (x, y, z) .$
Commented by Rasheed.Sindhi last updated on 12/Feb/22

$24!!!!!!!!!!!!!!!!! . . . I {couldn}^{'} t expect!$
Commented by Rasheed.Sindhi last updated on 12/Feb/22

$You can't use 'macro parameter character #' in math mode$
Commented by mr W last updated on 12/Feb/22

$4^{t h} d e g r e e \Rightarrow 4 r o o t s .$ $s y m m e t r y f o r x, y, z \Rightarrow e a c h r o o t s m e a n s 6 t r i p l e s$ $t h e r e f o r e 4 \times 6 = 24 t r i p l e s$
	Answered by mr W last updated on 12/Feb/22
	$S=(x+y)(y+z)(z+x) =2xyz+xy(x+y)+yz(y+z)+zx(z+x) =(x+y+z)(xy+yz+zx)−xyz =e_1 e_2 −e_3 p_2 =e_1 p_1 −2e_2 ⇒e_1 ^2 −2e_2 =70 p_3 =e_1 p_2 −e_2 p_1 +3e_3 ⇒70e_1 −e_1 e_2 +3e_3 =64 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 ⇒64e_1 −70e_2 +e_1 e_3 =2002 6e_3 =e_1 ^3 −210e_1 +128 e_1 ^4 −420e_1 ^2 +512e_1 +2688=0 (e_1 +2)(e_1 ^3 −2e_1 ^2 −416e_1 +1344)=0 ⇒e_1 =−2 ⇒e_1 ^3 −2e_1 ^2 −416e_1 +1344=0 let e_1 =s+(2/3) s^3 −((1252)/3)s+((28784)/(27))=0 s=((4(√(313)))/3) sin (((2kπ)/3)+(1/3) sin^(−1) ((1799)/( 313(√(313))))) ⇒e_1 =(2/3)+((4(√(313)))/3) sin (((2kπ)/3)+(1/3) sin^(−1) ((1799)/( 313(√(313))))) e_2 =(e_1 ^2 /2)−35 e_3 =(e_1 ^3 /6)−35e_1 +((64)/3) S=e_1 e_2 −e_3 =((e_1 ^3 −64)/3) ⇒S_1 =(((−2)^3 −64 )/3)=−24 ⇒S_(2,3,4) =(1/3){[(2/3)+((4(√(313)))/3) sin (((2kπ)/3)+(1/3) sin^(−1) ((1799)/( 313(√(313)))))]^3 −64} ≈−9.7517, 2516.6841, −3080.2657$
	$S = (x + y) (y + z) (z + x)$ $= 2 x y z + x y (x + y) + y z (y + z) + z x (z + x)$ $= (x + y + z) (x y + y z + z x) - x y z$ $= e_{1} e_{2} - e_{3}$ $p_{2} = e_{1} p_{1} - 2 e_{2}$ $\Rightarrow e_{1}^{2} - 2 e_{2} = 70$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$ $\Rightarrow 70 e_{1} - e_{1} e_{2} + 3 e_{3} = 64$ $p_{4} = e_{1} p_{3} - e_{2} p_{2} + e_{3} p_{1}$ $\Rightarrow 64 e_{1} - 70 e_{2} + e_{1} e_{3} = 2002$ $6 e_{3} = e_{1}^{3} - 210 e_{1} + 128$ $e_{1}^{4} - 420 e_{1}^{2} + 512 e_{1} + 2688 = 0$ $(e_{1} + 2) (e_{1}^{3} - 2 e_{1}^{2} - 416 e_{1} + 1344) = 0$ $\Rightarrow e_{1} = - 2$ $\Rightarrow e_{1}^{3} - 2 e_{1}^{2} - 416 e_{1} + 1344 = 0$ $l e t e_{1} = s + \frac{2}{3}$ $s^{3} - \frac{1252}{3} s + \frac{28784}{27} = 0$ $s = \frac{4 \sqrt{313}}{3} \sin (\frac{2 k π}{3} + \frac{1}{3} \sin^{- 1} \frac{1799}{313 \sqrt{313}})$ $\Rightarrow e_{1} = \frac{2}{3} + \frac{4 \sqrt{313}}{3} \sin (\frac{2 k π}{3} + \frac{1}{3} \sin^{- 1} \frac{1799}{313 \sqrt{313}})$ $e_{2} = \frac{e_{1}^{2}}{2} - 35$ $e_{3} = \frac{e_{1}^{3}}{6} - 35 e_{1} + \frac{64}{3}$ $S = e_{1} e_{2} - e_{3} = \frac{e_{1}^{3} - 64}{3}$ $\Rightarrow S_{1} = \frac{{(- 2)}^{3} - 64}{3} = - 24$ $\Rightarrow S_{2, 3, 4} = \frac{1}{3} {{[\frac{2}{3} + \frac{4 \sqrt{313}}{3} \sin (\frac{2 k π}{3} + \frac{1}{3} \sin^{- 1} \frac{1799}{313 \sqrt{313}})]}^{3} - 64}$ $\approx - 9.7517, 2516.6841, - 3080.2657$
	Commented by Rasheed.Sindhi last updated on 12/Feb/22

	$G reat sir!$
	Commented by nurtani last updated on 12/Feb/22

	$G o o d j o b s i r!!!$
	Commented by Tawa11 last updated on 12/Feb/22

	$Great sir$
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