Question

Answer this question step by step, I don't want any invalid answer ​

Answer 1

Answer:

The correct answer is Option A: $(\frac{p}{q} )^{2p}$.

Step-by-step explanation:

= $\frac{(p + \frac{1}{q} )^{p-q} * (p - \frac{1}{q} )^{p+q} }{(q + \frac{1}{p} )^{p-q} * (q - \frac{1}{p} )^{p+q} }$

= $\frac{(\frac{p}{1} + \frac{1}{q} )^{p-q} * (\frac{p}{1} - \frac{1}{q} )^{p+q} }{(\frac{q}{1} + \frac{1}{p} )^{p-q} * (\frac{q}{1} - \frac{1}{p} )^{p+q} }$

= $\frac{(\frac{(pq+1)}{q} )^{p-q} * (\frac{(pq-1)}{q} )^{p+q} }{(\frac{q}{1} + \frac{1}{p} )^{p-q} * (\frac{q}{1} - \frac{1}{p} )^{p+q} }$

= $\frac{(\frac{(pq+1)}{q} )^{p-q} * (\frac{(pq-1)}{q} )^{p+q} }{(\frac{(pq+1)}{p} )^{p-q} * (\frac{(pq-1)}{p} )^{p+q} }$

= $\frac{(pq+1)^{p-q} }{(q)^{p-q} } * \frac{(pq-1)^{p+q} }{(q)^{p+q} }$ ÷ $\frac{(pq+1)^{p-q} }{(p)^{p-q} } * \frac{(pq-1)^{p+q} }{(p)^{p+q}}$

= $\frac{(pq+1)^{p-q} * (pq-1)^{p+q} }{(q)^{p-q+p+q} }$ ÷ $\frac{(pq+1)^{p-q} * (pq-1)^{p+q} }{(p)^{p-q+p+q} }$

=  $\frac{(pq+1)^{p-q} * (pq-1)^{p+q} }{(q)^{2p} }$ ÷ $\frac{(pq+1)^{p-q} * (pq-1)^{p+q} }{(p)^{2p} }$

= $\frac{(pq+1)^{p-q} * (pq-1)^{p+q} }{(q)^{2p} }$ × $\frac{(p)^{2p}}{(pq+1)^{p-q} * (pq-1)^{p+q} }$

= $\frac{(p)^{2p} }{(q)^{2p} }$

= $(\frac{p}{q} )^{2p}$