Question

Write in single exponent form:

${(-a)}^4\times {(-b)}^4$

Answer 1

In ${(-a)}^4\times {(-b)}^4$, there are different bases but the exponents are same.

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

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

${(-a)}^4\times {(-b)}^4$$=\left[\right(-a)\times (-a)\times (-a)\times (-a\left)\right]\times \left[\right(-b)\times (-b)\times (-b)\times (-b\left)\right]$ (Here $a=-a,b=-b,m=4$)

$=\left[\right(-a)\times (-b\left)\right]\times \left[\right(-a)\times (-b\left)\right]\times \left[\right(-a)\times (-b\left)\right]\times \left[\right(-a)\times (-b\left)\right]$

$={\left(-a\times -b\right)}^4$

$={\left(ab\right)}^4$ ($-\times -=+$)

Therefore, ${(-a)}^4\times {(-b)}^4$$={\left(ab\right)}^4$.