{ ((x+y+z = 7)),((x^2 +y^2 +z^2 = 49)),((x^3 +y^3 +z^3 = 7)) :} find x; y ; z


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Question Number 95106 by bobhans last updated on 23/May/20
${ ((x+y+z = 7)),((x^2 +y^2 +z^2 = 49)),((x^3 +y^3 +z^3 = 7)) :} find x; y ; z$
${\begin{cases} x + y + z = 7 \\ x^{2} + y^{2} + z^{2} = 49 \\ x^{3} + y^{3} + z^{3} = 7 \end{cases}$ $f i n d x; y; z$
	Answered by john santu last updated on 23/May/20

	$(1) x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - (x y + x z + y z))$ $\Rightarrow 7 - 3 x y z = 7 (49 - (x y + x z + y z))$ $(2) {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + x z + y z)$ $\Rightarrow 49 = 49 + 2 (x y + x z + y z)$ $x y + x z + y z = 0$ $s o w e g e t 7 - 3 x y z = 7 \times 49$ $x y z = - 112$
	Commented by 1549442205 last updated on 23/May/20

	$That is not fully answer! I have the following answer :$ $We have 2 (xy + yz + zx) = {(x + y + z)}^{2} - (x^{2} + y^{2}$ $+ z^{2}) = 7^{2} - 49 = 0 \to xy + yz + zx = 0 \to yz +$ $(y + z) x = 0 \to yz + (7 - x) x = 0 \to yz = x^{2} - 7 x (1)$ $On the other hands, x^{3} + y^{3} + z^{3} = 7 \Leftrightarrow$ $x^{3} + [{(y + z)}^{3} - 3 yz (y + z)] = 7 \Leftrightarrow x^{3} + [{(7 - x)}^{3} -$ $3 (x^{2} - 7 x) (7 - x)] = 7 \Leftrightarrow x^{3} + [343 - 147 x +$ ${21 x}^{2} - x^{3} - 3 (- x^{3} + {14 x}^{2} - 49 x)] = 7 \Leftrightarrow$ ${3 x}^{3} - {21 x}^{2} + 336 = 0 \Leftrightarrow x^{3} - {7 x}^{2} + 112 = 0 .$ $Putting x = \frac{1}{t} we get$ ${112 t}^{3} - 7 t + 1 = 0 . Multiplying two sides of$ $the equation by 196 = 7^{2} . 4 and note 112 =$ $7 . 4^{2} then set u = 28 t we get PT : u^{3} - 49 u +$ $196 = 0 (1)$ $Put u = - (m + n) \to u + m + n = 0 . Using the$ $equality a^{3} + b^{3} + c^{3} - 3 abc = (a + b + c) (a^{2} +$ $b^{2} + c^{2} - ab - bc - ca) we have u^{3} + m^{3} + n^{3} -$ $3 umn = 0 (2) . From (1) and (2) we have :$ $m^{3} + n^{3} = 196 (3), mn = \frac{49}{3} . It follows$ ${(m^{3} - n^{3})}^{2} = {(m^{3} + n^{3})}^{2} - 4 {(mn)}^{3} = 196^{2} -$ $4 . {(\frac{49}{3})}^{3} = {(\frac{98}{9} \sqrt{177})}^{2} \Rightarrow m^{3} - n^{3} = \frac{98}{9} \sqrt{177} (2)$ $. Hence, from (1), (2) we get m^{3} = ($ $\frac{98 \sqrt{177}}{9} + 196) / 2 and n^{3} = (196 - \frac{98 \sqrt{177}}{9}) / 2$ $\Rightarrow u = - (m + n) =$ $- [(^{3} \sqrt{\frac{1764 + 98 \sqrt{177}}{18}} +^{3} \sqrt{\frac{1764 - 98 \sqrt{177}}{18}})$ $Since u = 28 tand t = \frac{1}{x} \Rightarrow x = \frac{28}{u} = - \frac{1}{7} [^{3} \sqrt{{(\frac{1764 + 98 \sqrt{177}}{18})}^{2}} - \frac{49}{3} +^{3} \sqrt{{(\frac{1764 - 98 \sqrt{177}}{18})}^{2}}]$ $Replace this value of x into the equations :$ $x + y + z = 7 and x (y + z) + yz = 0 we find y, z$ $which are image numbers$
	Commented by john santu last updated on 23/May/20

	$what your answer ?$
	Commented by bobhans last updated on 23/May/20

	$waw . . . . superb sir$
	Answered by mr W last updated on 23/May/20

	$e_{1} = p_{1} = x + y + z = 7$ $2 e_{2} = p_{1}^{2} - p_{2} = 7^{2} - 49 = 0 \Rightarrow x y + y z + z x = 0$ $3 e_{3} = \frac{1}{2} p_{1}^{3} - \frac{3}{2} p_{1} p_{2} + p_{3} = \frac{7^{3}}{2} - \frac{3 \times 7 \times 49}{2} + 7 = - 336 \Rightarrow x y z = - 112$ $x, y, z a r e r o o t s o f$ $t^{3} - 7 t^{2} + 112 = 0$ $. . .$
	Commented by bobhans last updated on 23/May/20

	$alright, the answer is very^{'} {smart}^{'}$
	Commented by bobhans last updated on 23/May/20

	$not finish ?$
	Commented by mr W last updated on 23/May/20

	$t h e r e s t i s j u s t ‘ ‘ {s t u p i d}^{″} c a l c u l a t i o n$ $w i t h v a l u e s . y o u c a n d o i t w h e n y o u$ $l i k e .$
	Commented by mr W last updated on 23/May/20

	$i t h i n k t h e m a i n p o i n t o f t h i s q u e s t i o n$ $i s t o g e t t h a t x, y, z a r e t h e t h r e e r o o t s$ $o f e q n . t^{3} - 7 t^{2} + 112 = 0 . t o c a l c u l a t e$ $t h e r o o t s (o n e r e a l a n d t w o c o m p l e x)$ $o f t h e c u b i c e q n . i s a h a r d b u t a l s o e a s y$ $d i l i g e n c e w o r k .$
	Commented by 1549442205 last updated on 23/May/20

	$This equation has unique real root$ $t = \frac{- 1}{7} [^{3} \sqrt{{(\frac{1764 + 98 \sqrt{177}}{18})}^{2}} - \frac{49}{3} +^{3} \sqrt{{(\frac{1764 - 98 \sqrt{177}}{18})}^{2}}] like I done above .$
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