Eliminating $x,y,z$ from given set of equalities 
Given that: $x-2y+z=a$, $x^2-2y^2+z^2=b$, $x^3-2y^3+z^3=c$ and $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0$. Eliminate $x,y,z$ from given set of equations. [Hint: Use $x^2+z^2=(x+z)^2-2xz$ and $x^3+z^3=(x+z)(x^2+z^2)-xz(x+z)$]

Now using hints as given, I can solve $y$ in terms of $a,b,c$, but now sure how to do it for $x,z$. So for example, for $y$:
$$\begin{aligned}b&=(x^2+z^2)-2y^2 \\ &=(x+z)^2-2xz-2y^2\\ &=(a+2y)^2-y(a+2y)-2y^2\end{aligned}$$
If we plug in the solution of $y$ from this into the third equation, one involving third powers of $x,y,z$, then we would be left with an expression only in $a,b,c$, but doing so is quite laborious. Is there any alternate method?
How to proceed? Any hints are appreciated. Thanks.

We leave it to the OP to show that if $a = 0$ then $b = 0$ and $c =0$ and $x=y=z$.
Else
$\tag 1 \Large{y = \frac{b-a^2}{3a}}$
and by the 4th equation $b \ne a^2$.
and since
$x^3+z^3=(x+z)(x^2+z^2)-xz(x+z)= (x+z)\big((x+z)^2-2xz\big)-xz(x+z)$
and
$xz = \frac{1}{2} (x+z)y$
we have
$x^3+z^3= (x+z)\big((x+z)^2-(x+z)y\big)-\frac{1}{2} (x+z)y(x+z)$
and therefore
$c + 2y^3 = (x+z)\big((x+z)^2-(x+z)y\big)-\frac{1}{2} (x+z)y(x+z)$
And so
$\tag 2 c + 2y^3 = (a+2y)^3 - \frac{3}{2} (a+2y)^2 y$

Work check
Let $a = 1$ and $b = 4$ so that $y = 1$.
Solving $\text{(2)}$ we get $c = 11.5$
Returning to the original system we find
$x = \frac{1}{2}(3 + \sqrt3) \land z = \frac{1}{2} (3 - \sqrt3)$
or
$x = \frac{1}{2}(3 - \sqrt3) \land z = \frac{1}{2} (3 + \sqrt3)$
