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Laws of Exponents for Real Numbers

Prove that ...

Question

Prove that $\frac{{(a^{p + q})}^{2} {(a^{q + r})}^{2} {(a^{r + p})}^{2}}{{(a^{p} . a^{q} . a^{r})}^{4}} = 1$

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Solution

$\frac{(a^{2 p + 2 q}) (a^{2 q + 2 r}) (a^{2 r + 2 p})}{a^{4 p} . a^{4 q} . a^{4 r}}$

$\Rightarrow \frac{a^{4 p + 4 q + 4 r}}{a^{4 p + 4 q + 4 r}}$

$= 1$

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