If a=xyp−1,b=xyq−1 and c=xyr−1,

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:

It is given that $a=x y^{p-1}, b=x y^{q-1}$ and $c=x y^{r-1}$.

$\therefore a^{q-r} b^{r-p} c^{p-q}$

$=\left(x y^{p-1}\right)^{q-r}\left(x y^{q-1}\right)^{r-p}\left(x y^{r-1}\right)^{p-q}$

$=x^{(q-r)} u^{(p-1)(q-r)} x^{(r-p)} u^{(r-p)(q-1)} x^{(p-q)} u^{(p-q)(r-1)}$

$=x^{(q-r)} x^{(r-p)} x^{(p-q)} y^{(p-1)(q-r)} y^{(r-p)(q-1)} y^{(p-q)(r-1)}$

$=x^{(q-r)+(r-p)+(p-q)} y^{(p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1)}$

$=x^{q-r+r-p+p-q} y^{p q-q-p r+r+r q-r-p q+p+p r-p-q r+q}$

$=x^{0} y^{0}$

$=1$

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