Question

If $a=x{y}^{p-1},b=x{y}^{q-1}\mathrm{and}c=x{y}^{r-1}$, prove that $a^{q-r}{b}^{r-p}{c}^{p-q}=1$.

Answer 1

It is given that $a=x{y}^{p-1},b=x{y}^{q-1}\mathrm{and}c=x{y}^{r-1}$.
$$\begin{gathered} \therefore a^{q-r}{b}^{r-p}{c}^{p-q} \\ ={\left(x{y}^{p-1}\right)}^{q-r}{\left(x{y}^{q-1}\right)}^{r-p}{\left(x{y}^{r-1}\right)}^{p-q} \\ =x^{\left(q-r\right)}{y}^{\left(p-1\right)\left(q-r\right)}{x}^{\left(r-p\right)}{y}^{\left(r-p\right)\left(q-1\right)}{x}^{\left(p-q\right)}{y}^{\left(p-q\right)\left(r-1\right)} \\ =x^{\left(q-r\right)}{x}^{\left(r-p\right)}{x}^{\left(p-q\right)}{y}^{\left(p-1\right)\left(q-r\right)}{y}^{\left(r-p\right)\left(q-1\right)}{y}^{\left(p-q\right)\left(r-1\right)} \\ =x^{\left(q-r\right)+\left(r-p\right)+\left(p-q\right)}{y}^{\left(p-1\right)\left(q-r\right)+\left(r-p\right)\left(q-1\right)+\left(p-q\right)\left(r-1\right)} \\ =x^{q-r+r-p+p-q}{y}^{pq-q-pr+r+rq-r-pq+p+pr-p-qr+q} \\ =x^0{y}^0 \\ =1 \end{gathered}$$