Question

I don’t understand it

Answer 1

Let's deal with each factor in the numerator individually first: when evaluating the power of a power, you have to multiply the exponents: $(a^b)^c = a^{bc}$

So, we have

$[tex](3^2)^{-\frac{1}{2}} = 3^{2\cdot(-\frac{1}{2})} = 3^{-1},\quad (9^4)^{-1} = 9^{-4}$

Now remember that $9=3^2$ and $27=3^3$ to rewrite the expression as

$\dfrac{3^{-1}\cdot (3^2)^{-4}}{(3^3)^{-3}} = \dfrac{3^{-1}\cdot 3^{-8}}{3^{-9}}$

Now use the rules

$a^b\cdot a^c = a^{b+c},\quad \dfrac{a^b}{a^c} = a^{b-c}$

to conclude

$\dfrac{3^{-1}\cdot 3^{-8}}{3^{-9}} = 3^{-1-8-(-9) = 3^0 = 1$