Solve -8(i)^3-6(1)^2+6(1)-9=

$$- 8 ( i ) ^ { 3 } - 6 ( 1 ) ^ { 2 } + 6 ( 1 ) - 9 =$$

Evaluate

$$-9+8i$$

Solution Steps

Calculate i to the power of 3 and get -i.

$$-8\left(-i\right)-6\times 1^{2}+6\times 1-9$$

Multiply -8 and -i to get 8i.

$$8i-6\times 1^{2}+6\times 1-9$$

Calculate 1 to the power of 2 and get 1.

$$8i-6\times 1+6\times 1-9$$

Multiply 6 and 1 to get 6.

$$8i-6+6\times 1-9$$

Do the additions.

$$6\times 1-15+8i$$

Multiply 6 and 1 to get 6.

$$6-15+8i$$

Do the additions.

$$-9+8i$$

Real Part

$$-9$$

Solution Steps

Calculate i to the power of 3 and get -i.

$$Re(-8\left(-i\right)-6\times 1^{2}+6\times 1-9)$$

Multiply -8 and -i to get 8i.

$$Re(8i-6\times 1^{2}+6\times 1-9)$$

Calculate 1 to the power of 2 and get 1.

$$Re(8i-6\times 1+6\times 1-9)$$

Multiply 6 and 1 to get 6.

$$Re(8i-6+6\times 1-9)$$

Do the additions in 8i-6-9.

$$Re(6\times 1-15+8i)$$

Multiply 6 and 1 to get 6.

$$Re(6-15+8i)$$

Do the additions in 6-15+8i.

$$Re(-9+8i)$$

The real part of -9+8i is -9.

$$-9$$