Writing $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\frac1x+\frac1y+\frac1z$

Question

Can we write $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\dfrac1x+\dfrac1y+\dfrac1z$?

What about $x^3+y^3+z^3$?

Answer 1

Let our three expressions be $a$, $b$, and $c$. Note that $xy+yz+zx=ac$.

For $x^2+y^2+z^2$, use the identity $$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz.$$

For $x^3+y^3+z^3$, use the not difficult to verify identity $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz).$$

Answer 2

Yes, by the Fundamental Theorem of Symmetric Polynomials since $x^2+y^2+z^2$ is symmetric and $xy+yz+xz=xyz*(1/x+1/y+1/z).$

Answer 3

$$x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx) = (x+y+z)^2 - 2(xyz)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$

$$x^3+y^3+z^3 = (x+y+z)^3 - (3x^2y + 3x^2z + 3xy^2 + 3y^2z + 3xz^2 + 3yz^2 + 6xyz) = (x+y+z)^3 - (3xy(x+y+z) + 3yz(x+y+z) + 3zx(x+y+z) - 3xyz)$$

$$= (x+y+z)^3 - 3(x+y+z)(xy+yz+zx) + 3xyz$$

$$= (x+y+z)^3 - 3(x+y+z)(xyz)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) + 3xyz$$