Solve 91-4^2+(3-5)*(-2)^3+3^2-(-2)^4

$$91 - 4 ^ { 2 } + ( 3 - 5 ) \cdot ( - 2 ) ^ { 3 } + 3 ^ { 2 } - ( - 2 ) ^ { 4 }$$

Evaluate

$$84$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$91-16+\left(3-5\right)\left(-2\right)^{3}+3^{2}-\left(-2\right)^{4}$$

Subtract 16 from 91 to get 75.

$$75+\left(3-5\right)\left(-2\right)^{3}+3^{2}-\left(-2\right)^{4}$$

Subtract 5 from 3 to get -2.

$$75-2\left(-2\right)^{3}+3^{2}-\left(-2\right)^{4}$$

Calculate -2 to the power of 3 and get -8.

$$75-2\left(-8\right)+3^{2}-\left(-2\right)^{4}$$

Multiply -2 and -8 to get 16.

$$75+16+3^{2}-\left(-2\right)^{4}$$

Add 75 and 16 to get 91.

$$91+3^{2}-\left(-2\right)^{4}$$

Calculate 3 to the power of 2 and get 9.

$$91+9-\left(-2\right)^{4}$$

Add 91 and 9 to get 100.

$$100-\left(-2\right)^{4}$$

Calculate -2 to the power of 4 and get 16.

$$100-16$$

Subtract 16 from 100 to get 84.

$$84$$

Factor

$$2^{2}\times 3\times 7$$