Where did you get it from?

We have,

x+y+z=3 ...................(1)

x^2+y^2+z^2=25......(2)

x^4+y^4+z^4=209....(3)

We shall use the identity,

(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc) (*)

Square equation (1) to get by (*),

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

Substitute (2) to get,

25+2(xy+xz+yz)=9

Thus,

xy+xz+yz=-8..............(4)

Square equation (2) to get by (*),

(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)=625

Substitute (3) to get,

209+2(x^2y^2+x^2z^2+y^2z^2)=625

Thus,

x^2y^2+x^2z^2+y^2z^2=208...(5)

Square equation (4) to get by (*),

(x^2y^2+x^2z^2+y^2z^2)+2(x^2yz+xy^2z+xyz^2)=64

Substitute (4) to get,

208+2(x^2yz+xy^2z+xyz^2)=64

208+2xyz(x+y+z)=64

Substitute (1) to get,

208+6xyz=64

Thus,

xyz=-24

Thus, we have 3 equations,

x+y+z=3

xy+xz+yz=-8

xyz=-24

To solve these equations we shall use Viete's theorem.

That means the solutions to the polynomial,

t^3-3t^2-8t+24=0

Are the solutions of those three equations.

Factor,

t^2(t-3)-8(t-3)=0

(t-3)(t^2-8)=0

Thus,

t=3 and t=sqrt(8) and t=-sqrt(8)

Thus, one solution without lose of generality is,

x=3

y=sqrt(8)

z=-sqrt(8)

Thus,

x^100+y^100+z^100=3^100+8^50+8^50=1.42724769270595 98810582859694495e+95