Find the Exact Value (2^3(3^2-2^0))/(2^3-3^2)

$\frac{2^{3} (3^{2} - 2^{0})}{(2^{3} - 3^{2})}$

Simplify the numerator.

Raise $3$ to the power of $2$ .

$\frac{2^{3} (9 - 2^{0})}{2^{3} - 3^{2}}$

Anything raised to $0$ is $1$ .

$\frac{2^{3} (9 - 1 \cdot 1)}{2^{3} - 3^{2}}$

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $9$ .

$\frac{2^{3} \cdot 8}{2^{3} - 3^{2}}$

Raise $2$ to the power of $3$ .

$\frac{8 \cdot 8}{2^{3} - 3^{2}}$

$\frac{8 \cdot 8}{2^{3} - 3^{2}}$

Simplify the denominator.

Raise $2$ to the power of $3$ .

$\frac{8 \cdot 8}{8 - 3^{2}}$

Raise $3$ to the power of $2$ .

$\frac{8 \cdot 8}{8 - 1 \cdot 9}$

Multiply $- 1$ by $9$ .

$\frac{8 \cdot 8}{8 - 9}$

Subtract $9$ from $8$ .

$\frac{8 \cdot 8}{- 1}$

$\frac{8 \cdot 8}{- 1}$

Simplify the expression.

Multiply $8$ by $8$ .

$\frac{64}{- 1}$

Divide $64$ by $- 1$ .

$- 64$

$- 64$

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