Question

The simplest form of ${27}^2\times 9^{-3}?$ is .

Answer 1

${27}^2\times 9^{-3}$
$=(3^3{)}^2\times (3^2{)}^{-3}$
[As $27=3\times 3\times 3=3^3$ and $9=3\times 3=3^2$]

$=3^{3\times 2}\times 3^{2\times (-3)}$ [As $(a^m{)}^n=a^{m\times n}$]
$=3^6\times 3^{-6}$
[ Multiplication of exponents with same base: $a^m\times a^n=a^{m+n}$ ]

$=3^{6+(-6)}$
$=3^{(6-6)}$
$=3^0=1$
[ As $a^0=1$ for all real $a\ne 0$]

Therefore, the simplest form of ${27}^2\times 9^{-3}$ is $1.$