Question

X+y+z=1, x^2+y^2+z^2=34 and x^3+y^3+z^3=97.solve x,yandz.

Answer 1

Answer:

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

Step-by-step explanation:

That happens since the power sums pk=xk+yk+zk for k=1,2,3 give you the values of the elementary symmetric functions e1=x+y+z,e2=xy+xz+yz,e3=xyz through Newton's identities. Then x,y,z can be identified with the roots of the polynomial:

p(w)=w3−e1w2+e2w−e3.

If you know in advance that (x,y,z)=(−1,−3,5) works, then every solution is given by a permutation of {−1,−3,5}, since the coefficients of p(w) are always the same, as well as its roots:

p(w)=w3−w2−17w−15=(w+1)(w+3)(w−5).

First note that

x2+y2+z2=(x+y+z)2−2(xy+yz+zx)=35

Now substituting x+y+z=1, we have 12−2(xy+yz+zx)=35, and thus xy+yz+zx=−17.

Then,

x3+y3+z3=(x+y+z)3−3(x+y+z)(xy+yz+zx)+3xyz=97

and substituting with 1 and our above result once again leads us to have xyz=15. Now we need a product of 15 and sum of 1 for (x,y,z), giving us (−1,−3,5) as a solution, amongst others.