Solve (3*2)^{2}:6^2+(2^2)^2:8-3^1=

$$( 3 \cdot 2 ) ^ { 2 } : 6 ^ { 2 } + ( 2 ^ { 2 } ) ^ { 2 } : 8 - 3 ^ { 1 } =$$

Evaluate

$$0$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\left(3\times 2\right)^{2}}{6^{2}}+\frac{2^{4}}{8}-3^{1}$$

Multiply 3 and 2 to get 6.

$$\frac{6^{2}}{6^{2}}+\frac{2^{4}}{8}-3^{1}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{36}{6^{2}}+\frac{2^{4}}{8}-3^{1}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{36}{36}+\frac{2^{4}}{8}-3^{1}$$

Divide 36 by 36 to get 1.

$$1+\frac{2^{4}}{8}-3^{1}$$

Calculate 2 to the power of 4 and get 16.

$$1+\frac{16}{8}-3^{1}$$

Divide 16 by 8 to get 2.

$$1+2-3^{1}$$

Add 1 and 2 to get 3.

$$3-3^{1}$$

Calculate 3 to the power of 1 and get 3.

$$3-3$$

Subtract 3 from 3 to get 0.

$$0$$

Factor

$$0$$