# If (y/z)^a * (z/x)^b*(x/y)^c = 1 Then prove (y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))?

May 8, 2018

#### Explanation:

As ${\left(\frac{y}{z}\right)}^{a} \cdot {\left(\frac{z}{x}\right)}^{b} \cdot {\left(\frac{x}{y}\right)}^{c} = 1$

${x}^{c - b} {y}^{a - c} {z}^{b - a} = 1$ .........(A)

Let ${\left(\frac{y}{z}\right)}^{\frac{1}{b - c}} = k$

then $y = {k}^{b - c} z$ and putting this in (A), we get

${x}^{c - b} {k}^{\left(b - c\right) \left(a - c\right)} {z}^{a - c} {z}^{b - a} = 1$

or ${x}^{c - b} {k}^{\left(b - c\right) \left(a - c\right)} {z}^{b - c} = 1$

or ${\left(\frac{z}{x}\right)}^{b - c} = {k}^{\left(b - c\right) \left(c - a\right)}$ (-note change from $a - c$ to $c - a$)

or $\frac{z}{x} = {k}^{c - a}$

or ${\left(\frac{z}{x}\right)}^{\frac{1}{c - a}} = k$

Similarly ${\left(\frac{x}{y}\right)}^{\frac{1}{a - b}} = k$

Hence ${\left(\frac{y}{z}\right)}^{\frac{1}{b - c}} = {\left(\frac{z}{x}\right)}^{\frac{1}{c - a}} = {\left(\frac{x}{y}\right)}^{\frac{1}{a - b}}$