x+y+z = 3

x^2+y^2+z^2= 9

xyz= -2

solve for x^4+y^4+z^4=?

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We have x+y+z=3

x^2+y^2+z^2=9

xyz=-2

Now

`(x+y+z)^2=(3)^2`

`x^2+y^2+z^2+2(xy+yz+zx)=9`

`=> 9+2(xy+yz+zx)=9`

`xy+yz+zx=0` (i)

`(xy+yz+zx)^2=(xy)^2+(yz)^2+(zx)^2+2(xy^2z+xyz^2+x^2yz)`

`0=(xy)^2+(yz)^2+(zx)^2+2xyz(x+y+z)`

`(xy)^2+(yz)^2+(zx)^2=-2xyz(x+y+z)``=-2(-2)(3)=12` (ii)

Thus

`(x^2+y^2+z^2)^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)`

`x^4+y^4+z^4=(x^2+y^2+z^2)^2-2((xy)^2+(yz)^2+(zx)^2)`

`=(9)^2-2xx12 `

`=69` using (ii).

We have used following

`(A+B+C)^2=A^2+B^2+C^2+2(AB+BC+CA)`

thanks! but you miscalculated

9^2 -2x12 = 81-24=57

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