Question:
If $a=x y^{p-1}, b=x y^{q-1}$ and $c=x y^{r-1}$, prove that $a^{q-r} b^{r-p} c^{p-q}=1$
Solution:
Given, $a=x y^{p-1}, b=x y^{q-1}$ and $c=x y^{r-1}$
To prove, $a^{q-r} b^{r-p} c^{p-q}=1$
Left hand side (LHS) = Right hand side (RHS)
Considering LHS, $=a^{q-r} b^{r-p} c^{p-q} \ldots \ldots$ (i)
By substituting the value of $a, b$ and $c$ in equation (i), we get
$=\left(x y^{p-1}\right)^{q-r}\left(x y^{q-1}\right)^{r-p}\left(x y^{r-1}\right)^{p-q}$
$=x y^{p q-p r-q+r} x y^{q r-p q-r+p} x y^{r p-r q-p+q}$
$=x y^{p q-p r-q+r+q r-p q-r+p+r p-r q-p+q}$
$=x y^{0}$
$=1$