Question

What is the value of the expression

Answer 1

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

OR (in more details):

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

Answer 2

So either you meant $(\frac{-3^2}{-3^3} )*\frac{2}{1}$, or you actually meant $(\frac{-3^2}{-3^3} )^2$ . I'll solve both.

$(\frac{-3^2}{-3^3} )*\frac{2}{1}$

Firstly, do the division. Since the numerator and the denominator have the same base, you can write it as -3^-1, which can be rewritten as -1/3. $-\frac{1}{3}*\frac{2}{1}$

Next, multiply the fractions, and your answer should be $-\frac{2}{3}$

$(\frac{-3^2}{-3^3} )^2$

Firstly, multiply square with the powers in the fraction to get $\frac{-3^4}{-3^6}$

Next, divide. Like before, they have the same base, so the answer can be written as -3^-2, which can be rewritten as 1/3^2 = 1/9.