Evaluate (9^2-2^6)- square root of 64+36÷3(4)

$(9^{2} - 2^{6}) - \sqrt{64} + 36 \div 3 (4)$

Simplify each term.

Raise $9$ to the power of $2$ .

$81 - 2^{6} - \sqrt{64} + 36 \div 3 \cdot 4$

Raise $2$ to the power of $6$ .

$81 - 1 \cdot 64 - \sqrt{64} + 36 \div 3 \cdot 4$

Multiply $- 1$ by $64$ .

$81 - 64 - \sqrt{64} + 36 \div 3 \cdot 4$

Rewrite $64$ as $8^{2}$ .

$81 - 64 - \sqrt{8^{2}} + 36 \div 3 \cdot 4$

Pull terms out from under the radical, assuming positive real numbers.

$81 - 64 - 1 \cdot 8 + 36 \div 3 \cdot 4$

Multiply $- 1$ by $8$ .

$81 - 64 - 8 + 36 \div 3 \cdot 4$

Divide $36$ by $3$ .

$81 - 64 - 8 + 12 \cdot 4$

Multiply $12$ by $4$ .

$81 - 64 - 8 + 48$

Simplify by adding and subtracting.

Subtract $64$ from $81$ .

$17 - 8 + 48$

Subtract $8$ from $17$ .

$9 + 48$

Add $9$ and $48$ .

$57$