Question

x + y + z = 1
x2 + y2 + z2 = 3
x3 + y3 + z3 = 7
Find. x5 + y5 + z5
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Answer 1

Step-by-step explanation:

x+y+z=1

x2+y2+z2=2

x3+y3+z3=3

Then find the value of

x5+y5+z5

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The problem at hand is a special case of 3 variables. Let e1,e2,e3 be the three elementary symmetric polynomial associated with x,y,z:

e1e2e3=x+y+z=xy+yz+zx=xyz

This is equivalent to

(λ−x)(λ−y)(λ−z)=λ3−e1λ2+e2λ−e3.

For any integer k>0, let pk=xk+yk+zk. For 3 variables, the Newton's identities are:

p1p2p3pn=e1=e1p1−2e2=e1p2−e2p1+3e3=e1pn−1−e2pn−2+e3pn−3 for n>3

Given the known values of (p1,p2,p3)=(1,2,3), we have

e1e2e3=p1=1;=−12(p2−e1p1)=−12(2−12)=−12=13(p3−e1p2+e2p1)=13(3−2−12)=16

Substitute this into formula of p4 and p5, we have

p4p5=e1p3−e2p2+e3p1=3+12⋅2+16⋅1=256=e1p4−e2p3+e3p2=256+12⋅3+16⋅2=6

The answer is x5+y5+z5=6.

Answer 2

Answer:

$x {}^{5} + y {}^{5} + z {}^{5} = 12.$