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Question
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`
Solution
It is given that `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1)`
`thereforea^(q-r)b^(r-p)c^(p-q)`
`=(xy^(p-1))^(q-r)(xy^(q-1))^(r-p)(xy^(r-1))^(p-q)`
`=x^((q-r))y^((p-1)(q-r))x^((r-p))y^((r-p)(q-1))x^((p-q))y^((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^(pq-q-pr+r+rq-r-pq+p+pr-p-qr+q)`
`=x^0y^0`
= 1
Hence proved.
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