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Question Number 165984 by nurtani last updated on 10/Feb/22

Commented by cortano1 last updated on 10/Feb/22

$2024$
	Answered by Rasheed.Sindhi last updated on 11/Feb/22
	${ ((x^2 +y^2 +z^2 =70........(i))),((x^3 +y^3 +z^3 =64........(ii))),(((x+y)(y+z)(z+x)=−24....(iii))),((x^4 +y^4 +z^4 +2=?)) :} •(x+y+z)^3 =x^3 +y^3 +z^3 +3(x+y)(y+z)(z+x) (x+y+z)^3 =64+3(−24)=−8 determinant ((( x+y+z=−2))) • (x+y+z)^2 =(−2)^2 x^2 +y^2 +z^2 +2(xy+yz+zx)=4 70+2(xy+yz+zx)=4 determinant (((xy+yz+zx=−33))) •x^3 +y^3 +z^3 −3xyz=(x+y+z)(x^2 +y^2 +z^2 −xy−yz−zx) 64−3xyz=(−2)(70−(−33)) xyz=((64+206)/3)=((270)/3)=90 determinant (((xyz=90))) •(xy+yz+zx)^2 =(−33)^2 x^2 y^2 +y^2 z^2 +z^2 x^2 +2xyz(x+y+z)=1089 x^2 y^2 +y^2 z^2 +z^2 x^2 +2(90)(−2)=1089 x^2 y^2 +y^2 z^2 +z^2 x^2 =1089+360=1449 determinant (((x^2 y^2 +y^2 z^2 +z^2 x^2 =1449))) •(x^2 +y^2 +z^2 )^2 =x^4 +y^4 +z^4 +2(x^2 y^2 +y^2 z^2 +z^2 x^2 ) (70)^2 =x^4 +y^4 +z^4 +2(1449) x^4 +y^4 +z^4 =2002 x^4 +y^4 +z^4 +2=2004 determinant (((x^4 +y^4 +z^4 +2=2004)))$
	${\begin{cases} x^{2} + y^{2} + z^{2} = 70 . . . . . . . . (i) \\ x^{3} + y^{3} + z^{3} = 64 . . . . . . . . (i i) \\ (x + y) (y + z) (z + x) = - 24 . . . . (i i i) \\ x^{4} + y^{4} + z^{4} + 2 = ? \end{cases}$ $∙ {(x + y + z)}^{3} = x^{3} + y^{3} + z^{3} + 3 (x + y) (y + z) (z + x)$ ${(x + y + z)}^{3} = 64 + 3 (- 24) = - 8$ $\begin{matrix} x + y + z = - 2 \end{matrix}$ $∙ {(x + y + z)}^{2} = {(- 2)}^{2}$ $x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x) = 4$ $70 + 2 (x y + y z + z x) = 4$ $\begin{matrix} x y + y z + z x = - 33 \end{matrix}$ $∙ x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ $64 - 3 x y z = (- 2) (70 - (- 33))$ $x y z = \frac{64 + 206}{3} = \frac{270}{3} = 90$ $\begin{matrix} x y z = 90 \end{matrix}$ $∙ {(x y + y z + z x)}^{2} = {(- 33)}^{2}$ $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} + 2 x y z (x + y + z) = 1089$ $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} + 2 (90) (- 2) = 1089$ $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = 1089 + 360 = 1449$ $\begin{matrix} x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = 1449 \end{matrix}$ $∙ {(x^{2} + y^{2} + z^{2})}^{2} = x^{4} + y^{4} + z^{4} + 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ ${(70)}^{2} = x^{4} + y^{4} + z^{4} + 2 (1449)$ $x^{4} + y^{4} + z^{4} = 2002$ $x^{4} + y^{4} + z^{4} + 2 = 2004$ $\begin{matrix} x^{4} + y^{4} + z^{4} + 2 = 2004 \end{matrix}$
	Commented by Rasheed.Sindhi last updated on 11/Feb/22

	$T han k s for rapid feedback sir!$ $(Here some persons even {don}^{'} t give$ $any feedback .)$
	Commented by nurtani last updated on 11/Feb/22

	$N i c e, T h a n k y o u s i r$
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