How to deal with cubic root equations If x,y,z are real positive number, and the conditions are:
\begin{cases}
\begin{array}{ll}
1995x^3=1996y^3 \\
1996y^3=1997z^3 \\
\sqrt[3]{1995x^2+1996y^2+1997z^2}=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}
\end{array}
\end{cases}
What's the result of: 
\begin{equation}
\begin{array}{ll}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
\end{array}
\end{equation}
I can do below transfer,but don't know how to get rid of the x:
\begin{equation}
\begin{array}{ll}
\sqrt[3]{\frac{1995x^3}{x}+\frac{1996y^3}{y}+\frac{1997z^3}{z}}=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} \\
\sqrt[3]{1995x^3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}=\sqrt[3]{1995}+\sqrt[3]{1996} +\sqrt[3]{1997} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{(\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997})^3}{1995x^3}
\end{array}
\end{equation}
 A: Call $1995x^3=A$, $1996y^3=B$ and $1997z^3=C$.
Then $A=B=C$ and 
$(1995^{1/3}+1996^{1/3}+1997^{1/3})^3=1995\frac{A^{2/3}}{1995^{2/3}}+1996\frac{B^{2/3}}{1996^{2/3}}+1997\frac{C^{2/3}}{1997^{2/3}}=A^{2/3}(1995/1995^{2/3}+1997/1997^{2/3}+1997/1997^{2/3})$
Solve for $A=(\frac{(1995^{1/3}+1996^{1/3}+1997^{1/3})^3}{1995/1995^{2/3}+1997/1997^{2/3}+1997/1997^{2/3}})^{3/2}$ and you got it.
A: Call $1995x^3=1996y^3=1997z^3=M$,then
$$
\begin{cases}
\begin{align}
1995=\frac{M}{x^3} \tag{1} \\
1996=\frac{M}{y^3} \tag{2} \\
1997=\frac{M}{z^3} \tag{3}
\end{align}
\end{cases}
$$
So:
$$
\begin{align}
 & \sqrt[3]{1995x^2+1996y^2+1997z^2} =\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} \tag{4}\\
=> & \sqrt[3]{\frac{M}{x}+\frac{M}{y}+\frac{M}{z}} =\sqrt[3]{\frac{M}{x^3}}+\sqrt[3]{\frac{M}{y^3}}+\sqrt[3]{\frac{M}{z^3}} \tag{5} \\
=> & \sqrt[3]{M}\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} =\sqrt[3]{M}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \tag{6} \\
=> & (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 =1 \tag{7} \\
=> & \frac{1}{x}+\frac{1}{y}+\frac{1}{z} =1 \tag{8}
\end{align}
$$
