x+y+z = 3 x^2+y^2+z^2= 9 xyz= -2 solve for   x^4+y^4+z^4=?

2 Answers | Add Yours

kibauu's profile pic

Posted on

thanks! but you miscalculated

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

 

pramodpandey's profile pic

Posted on

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)`

We’ve answered 327,617 questions. We can answer yours, too.

Ask a question