Find all the complex solutions to the equation: \$z^3 + 8 = 0\$ .

Given equation

\$ z^3 + 2^3 \$

Factorizing the equation

\$ (z)^3 + (2)^3 = (z + 2)(z^2 - 2z + 4) \$

Setting each factor to zero

z + 2 = 0

z = - 2

Now, we set other factor to zero and solve for z:

\$ z^2 - 2z + 4 = 0 \$

For this quadratic equation, we can use the quadratic formula:

\$ z = ((-b) \pm \sqrt(b^2 - 4ac))/(2a) \$

Where, a = 1 , b = - 2 and c = 4.

Now plug the values in the formula

\$ z = ((-(-2)) \pm \sqrt((-2)^2 - 4(1)(4)))/(2(1)) \$

\$ z = (2 \pm \sqrt(- 12))/(2) \$

Simplify terms with \$i^2\$ (where \$i^2 = -1\$), then simplify

\$ z = (2 \pm 2i\sqrt(3))/(2) \$

Factor out 2 from the numerator and cancel it out to simplify the expression

\$ z = 1 \pm i\sqrt(3) \$

So, the three complex solutions to the equation \$ z^3 + 8 = 0\$ are \$ z = - 2 , z = 1 \pm i\sqrt(3) \$.

This is incorrect because it contains z = 1, but the correct root is z = −2

\$z = 1, z = (-1 \pm i sqrt(3)) \$

So, the correct option B includes all the solutions

\$ z = - 2 , z = 1 \pm i\sqrt(3) \$

This is wrong because it provides solutions of z = 2 and stem;[z = 2 ± i \sqrt(6)], which are not the roots of the given equation

\$z = 2, z = (2 \pm i sqrt(6)) \$

This is wrong because it doesn’t include the solution z = −2

\$z = 1, z = (1 \pm i sqrt(2)) \$

Option

B

Can you summarize what you’ve understood in the above steps?