Solve 7^{2}+(2^3)^2-3^2*2^2=

$$7 ^ { 2 } + ( 2 ^ { 3 } ) ^ { 2 } - 3 ^ { 2 } \cdot 2 ^ { 2 } =$$

Evaluate

$$77$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$7^{2}+2^{6}-3^{2}\times 2^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49+2^{6}-3^{2}\times 2^{2}$$

Calculate 2 to the power of 6 and get 64.

$$49+64-3^{2}\times 2^{2}$$

Add 49 and 64 to get 113.

$$113-3^{2}\times 2^{2}$$

Calculate 3 to the power of 2 and get 9.

$$113-9\times 2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$113-9\times 4$$

Multiply 9 and 4 to get 36.

$$113-36$$

Subtract 36 from 113 to get 77.

$$77$$

Factor

$$7\times 11$$