Question

Write in single exponent form:

${(-p)}^8\times {(-q)}^8$

Answer 1

In ${(-p)}^8\times {(-q)}^8$, there are different bases but the exponents are same.

So, for any non-zero integers $a,b$ where $m$ is any whole number,

$a^m\times b^m={\left(ab\right)}^m$

${(-p)}^8\times {(-q)}^8$$=\left[\right(-p)\times (-p)\times (-p)\times (-p)\times (-p)\times (-p)\times (-p)\times (-p\left)\right]\times \left[\right(-q)\times (-q)\times (-q)\times (-q)\times (-q)\times (-q)\times (-q)\times (-q\left)\right]$

(Here $a=-p,b=-q,m=8$)

$=\left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]\times \left[\right(-p)\times (-q\left)\right]$

$={\left(-p\times -q\right)}^8$

$={\left(pq\right)}^8$ ($-\times -=+$)

Therefore, ${(-p)}^8\times {(-q)}^8$$={\left(pq\right)}^8$.