Evaluate ((3^2)/(3^-1))^3

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

Step 1

Move $3^{- 1}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

${(3^{2} \cdot 3)}^{3}$

Step 2

Multiply $3^{2}$ by $3$ by adding the exponents.

Multiply $3^{2}$ by $3$ .

Raise $3$ to the power of $1$ .

${(3^{2} \cdot 3^{1})}^{3}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(3^{2 + 1})}^{3}$

Add $2$ and $1$ .

${(3^{3})}^{3}$

Step 3

Multiply the exponents in ${(3^{3})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$3^{3 \cdot 3}$

Multiply $3$ by $3$ .

$3^{9}$

Step 4

Raise $3$ to the power of $9$ .

$19683$