Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$

Given:

$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$

We have to show that :

$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$

I made three equations using cross multiplication :

$$1.~~x^{c-a}=y^{b-c}$$ $$2.~~y^{a-b}=z^{c-a}$$ $$3.~~z^{b-c}=x^{a-b}$$

How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.

Thank you.

We have $$\dfrac{\log(x)}{b-c} = \dfrac{\log(y)}{c-a} = \dfrac{\log(z)}{a-b} = t$$ This gives us $$x=e^{t(b-c)}, y = e^{t(c-a)} \text{ and }z = e^{t(a-b)}$$ Hence, \begin{align} x^{b+c-a}\cdot y^{c+a-b} \cdot z^{a+b-c} & = e^{t\left((b-c)(b+c-a) + (c-a)(c+a-b) + (a-b)(a+b-c)\right)}\\ & = e^{t(b^2-c^2-ab+ac + c^2 - a^2 -bc + ba + a^2 - b^2 - ac + bc)} = e^0 = 1 \end{align}

If you want to use your equations, here is a method.

Multiplying the equations together, we obtain: $$x^{c-a}y^{a-b}z^{b-c}=y^{b-c}z^{c-a}x^{a-b}$$ which gives after reordering: $$x^{b+c-a}y^{c+a-b}z^{a+b-c}=x^a y^b z^c$$ Therefore it suffices to show that $x^a y^b z^c = 1$.

Your first and third equations give $y = x^{\frac {c-a}{b-c}}, z = x^{\frac{a-b}{b-c}}$. This gives us: $$x^a y^b z^c = x^a x^{\frac {c-a}{b-c}\times b} x^{\frac{a-b}{b-c}\times c} = x^{a + \frac{bc-ba+ca-bc}{b-c}} = x^{a-a} = x^0 = 1 $$

QED.

Given: $$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}=\lambda$$ we have: $$ x = e^{\lambda(b-c)},\quad y=e^{\lambda(c-a)},\quad z=e^{\lambda(a-b)}, $$ hence: $$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = \exp\left(\lambda\cdot\sum_{cyc}\left(b^2-c^2-a(b-c)\right)\right)=\exp(0)=1.$$