Question

solve for z: z³-8=0 write your answers in simplified, rationlized form.

Answer 1

Answer:$\begin{gathered} z_{1\text{ }}=2 \\ z_2=\text{ -1 + i}\sqrt[]{3} \\ z_3=\text{ -1 - i}\sqrt[]{3} \end{gathered}$

Explanations:

The given equation is:

$z^3\text{ - 8 = 0}$

This can be rewritten as:

$z^3-2^3\text{ = 0}$

Note that:

$a^3-b^3=(a-b)(a^2\text{ + }ab+b^2)$

Therefore:

$z^3-2^{3\text{ }}=(z-2)(z^2\text{ + 2z + 4)}$$\begin{gathered} z\text{ - 2 = 0} \\ z_1\text{ = 2} \\ z^2\text{ + 2z + 4 = 0} \\ \text{The almighty formula can be used to solve the quadratic equation } \\ z\text{ = }\frac{\text{-b }\pm\sqrt[]{b^2-4ac}}{2a} \\ a\text{ = 1, b = 2, c = 4} \\ z\text{ = }\frac{\text{-2 }\pm\sqrt[]{2^2-4(1)(4)}}{2(1)} \\ z\text{ = }\frac{\text{-2 }\pm\sqrt[]{-12}}{2} \\ z\text{ = }\frac{\text{-2 }\pm i\sqrt[]{12}}{2} \\ z\text{ = }\frac{-2}{2}\pm\frac{i2\sqrt[]{3}}{2} \\ z\text{ = -1 }\pm i\sqrt[]{3} \\ z_2=\text{ -1+i}\sqrt[]{3} \\ z_3=\text{ -1 - i}\sqrt[]{3} \end{gathered}$