# Thread: find x^100 +y^100 +z^100

1. ## find x^100 +y^100 +z^100

x,y,z real numbers

x+y+z=3
x^2+y^2+z^2=25
x^4+y^4 +z^4=209
find
x^100 +y^100 +z^100

2. 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

3. Originally Posted by ThePerfectHacker
Where did you get it from?

...3^100+8^50+8^50=1.42724769270595988105828596944 95e+95
Hello, TPH,

I'm just curious where you got this number from?

When I used my calculator the result is:

3^100+8^50+8^50 = 518232016117423250798577701704520262974873015249 = 5.1823.. 10^47

EB

4. Originally Posted by earboth
Hello, TPH,

I'm just curious where you got this number from?
I am one of those calculating prodigies.
However, I realized my superior skills fails when numbers get too large. Thus, perhaps I made a mistake.