If ${(\frac{y}{z})}^{a} \cdot {(\frac{z}{x})}^{b} \cdot {(\frac{x}{y})}^{c} = 1$ $(y/z)^a * (z/x)^b*(x/y)^c = 1$ Then prove ${(\frac{y}{z})}^{\frac{1}{b - c}} = {(\frac{z}{x})}^{\frac{1}{c - a}} = {(\frac{x}{y})}^{\frac{1}{a - b}}$ $(y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))$ ?

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If ${(\frac{y}{z})}^{a} \cdot {(\frac{z}{x})}^{b} \cdot {(\frac{x}{y})}^{c} = 1$ $(y/z)^a * (z/x)^b*(x/y)^c = 1$ Then prove ${(\frac{y}{z})}^{\frac{1}{b - c}} = {(\frac{z}{x})}^{\frac{1}{c - a}} = {(\frac{x}{y})}^{\frac{1}{a - b}}$ $(y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))$ ?

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Answer 1 · 2018-05-08T11:20:51

Please see below.

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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If ${(\frac{y}{z})}^{a} \cdot {(\frac{z}{x})}^{b} \cdot {(\frac{x}{y})}^{c} = 1$ $(y/z)^a * (z/x)^b*(x/y)^c = 1$ Then prove ${(\frac{y}{z})}^{\frac{1}{b - c}} = {(\frac{z}{x})}^{\frac{1}{c - a}} = {(\frac{x}{y})}^{\frac{1}{a - b}}$ $(y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))$ ?

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If ${(\frac{y}{z})}^{a} \cdot {(\frac{z}{x})}^{b} \cdot {(\frac{x}{y})}^{c} = 1$ $(y/z)^a * (z/x)^b*(x/y)^c = 1$ Then prove ${(\frac{y}{z})}^{\frac{1}{b - c}} = {(\frac{z}{x})}^{\frac{1}{c - a}} = {(\frac{x}{y})}^{\frac{1}{a - b}}$ $(y/z)^(1/(b-c))=(z/x)^(1/(c-a)) = (x/y)^(1/(a-b))$ ?