2

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}

lucky1928
  • 159
  • What is the source of this problem? It seems bizarre and arbitrary. That said, you have three equations in three unknowns. It looks like it should be tedious, but not difficult, to sort it out. – lulu Aug 04 '17 at 21:49
  • Such a rarity, as literals suggest. :-} – bipll Aug 04 '17 at 22:32

2 Answers2

3

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.

Marja
  • 407
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: $$ \begin{align} & \sqrt[3]{1995x^2+1996y^2+1997z^2} =\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} \tag{4}\\ => & \sqrt[3]{\frac{M}{x}+\frac{M}{y}+\frac{M}{z}} =\sqrt[3]{\frac{M}{x^3}}+\sqrt[3]{\frac{M}{y^3}}+\sqrt[3]{\frac{M}{z^3}} \tag{5} \\ => & \sqrt[3]{M}\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} =\sqrt[3]{M}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \tag{6} \\ => & (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 =1 \tag{7} \\ => & \frac{1}{x}+\frac{1}{y}+\frac{1}{z} =1 \tag{8} \end{align} $$

lucky1928
  • 159
beetlej
  • 190