If $a = xy^{p-1}, b = xy^{q-1}$ and $c = xy^{r-1}$, prove that $a^{q-r} b^{r-p} c^{p-q} = 1$.

Given:

$a = xy^{p-1}, b = xy^{q-1}$ and $c = xy^{r-1}$

To do:

We have to prove that $a^{q-r} b^{r-p} c^{p-q} = 1$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=a^{q-r} b^{r-p} c^{p-q}$

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

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

$=x^{q-r+r-p+p-q} \times y^{p q-p r-q-r} \times y^{qr-p q-r+p} \times y^{rp-rq-p+q}$

$=x^{0} \times y^{p q-p r-q+r+q r-p q-r+p+r p-r q-p+q}$

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

$=1 \times 1$

$=1$

$=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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