How to deal with cubic root equations

Question

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}

Answer 1

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.

Answer 2

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: $$ \begin{align} & \sqrt[3]{1995x^2+1996y^2+1997z^2} =\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} \tag{4}\\ => & \sqrt[3]{\frac{M}{x}+\frac{M}{y}+\frac{M}{z}} =\sqrt[3]{\frac{M}{x^3}}+\sqrt[3]{\frac{M}{y^3}}+\sqrt[3]{\frac{M}{z^3}} \tag{5} \\ => & \sqrt[3]{M}\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} =\sqrt[3]{M}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \tag{6} \\ => & (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 =1 \tag{7} \\ => & \frac{1}{x}+\frac{1}{y}+\frac{1}{z} =1 \tag{8} \end{align} $$