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Solve for x, y and z if: x + y + z = 1 ; x^2 + y^2 +z^2 = 35 and x^3 + y^3 + z^3 = 97. Thanks.

x + y + z = 1

(x + y + z)^2 = 1

x^2 + y^2 +z^2 + 2xy + 2xz + 2yz = 1

35 + 2xy +2xz + 2yz = 1

2xy +2xz + 2yz = -34

xy + xz + yz = -17

-17 = (1)(xy + xz + yz ) = (x + y + z)(xy + xz + yz )

-17 = 3xyz + yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2

(x + y + z)^3 = 1

x^3 + y^3 + z^3 + 3(yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2) + 6xyz = 1

97 + 3(yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2) + 6xyz = 1

3(yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2) + 6xyz = -96

(yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2) + 2xyz = -32

Subtract the following two equations

-17 = 3xyz + yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2

-32 = 2xyz + yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2

15 = xyz

We now have

x + y + z = 1

xy + xz + yz = -17

xyz = 15

So,

(t-x)(t-y)(t-z) = t^3 - t^2 - 17t - 15

The roots of t^3 - t^2 - 17t - 15 are 5 , -1 , -3

So any permutation of 5 , -1 , -3 may be assigned to x , y , z

This gives 6 possible solutions to the given 3 equations

x=5 y=-1 z=-3 ; x=5 y=-3 z=-1 ; x=-1 y=5 z=-3 ; x=-1 y=-3 z=5 ; x=-3 y=-1 z=5 ;x=-3 y=5 z=-1

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