Solve ((i)^4-9(1)^3-30(1)-72)

$$( ( i ) ^ { 4 } - 9 ( 1 ) ^ { 3 } - 30 ( 1 ) - 72 )$$

Evaluate

$$-110$$

Solution Steps

Calculate i to the power of 4 and get 1.

$$1-9\times 1^{3}-30\times 1-72$$

Calculate 1 to the power of 3 and get 1.

$$1-9\times 1-30\times 1-72$$

Multiply 9 and 1 to get 9.

$$1-9-30\times 1-72$$

Subtract 9 from 1 to get -8.

$$-8-30\times 1-72$$

Multiply 30 and 1 to get 30.

$$-8-30-72$$

Subtract 30 from -8 to get -38.

$$-38-72$$

Subtract 72 from -38 to get -110.

$$-110$$

Real Part

$$-110$$

Solution Steps

Calculate i to the power of 4 and get 1.

$$Re(1-9\times 1^{3}-30\times 1-72)$$

Calculate 1 to the power of 3 and get 1.

$$Re(1-9\times 1-30\times 1-72)$$

Multiply 9 and 1 to get 9.

$$Re(1-9-30\times 1-72)$$

Subtract 9 from 1 to get -8.

$$Re(-8-30\times 1-72)$$

Multiply 30 and 1 to get 30.

$$Re(-8-30-72)$$

Subtract 30 from -8 to get -38.

$$Re(-38-72)$$

Subtract 72 from -38 to get -110.

$$Re(-110)$$

The real part of -110 is -110.

$$-110$$