Question

$\frac{{\left(a^{p+q}\right)}^{15}\cdot {\left(a^{q+r}\right)}^{15}\cdot {\left(a^{r+p}\right)}^{15}}{{\left(a^{p+q+r}\right)}^{29}}=$___

• A. 0
• B. 1
• C. $a$
• D. $\left(a^{p+q+r}\right)$

Answer 1

Answer: (C)

$\frac{{\left(a^{p+q}\right)}^{15}{\left(a^{q+r}\right)}^{15}{\left(a^{r+p}\right)}^{15}}{{\left(a^{p+q+r}\right)}^{29}}$
$=\frac{{\left(a^{p+q+q+r+r+p}\right)}^{15}}{{\left(a^{p+q+r}\right)}^{29}}$
$=\frac{{\left(a^{p+q+r}\right)}^{30}}{{\left(a^{p+q+r}\right)}^{29}}$
$=a^{p+q+r}$