Questions: Find all complex number solutions to the equation. x^3 - 512 = 0 Choose the correct solution set below. A. 4(cos 60° + i sin 60°), 4(cos 180° + i sin 180°), 4(cos 300° + i sin 300°) B. 8(cos 0° + i sin 0°), 8(cos 120° + i sin 120°), 8(cos 240° + i sin 240°) C. 4(cos 0° + i sin 0°), 4(cos 120° + i sin 120°), 4(cos 240° + i sin 240°) D. 8(cos 60° + i sin 60°), 8(cos 180° + i sin 180°), 8(cos 300° + i sin 300°)

Find all complex number solutions to the equation.
x^3 - 512 = 0

Choose the correct solution set below.
A. 4(cos 60° + i sin 60°), 4(cos 180° + i sin 180°), 4(cos 300° + i sin 300°)
B. 8(cos 0° + i sin 0°), 8(cos 120° + i sin 120°), 8(cos 240° + i sin 240°)
C. 4(cos 0° + i sin 0°), 4(cos 120° + i sin 120°), 4(cos 240° + i sin 240°)
D. 8(cos 60° + i sin 60°), 8(cos 180° + i sin 180°), 8(cos 300° + i sin 300°)

Transcript text: Find all complex number solutions to the equation. \[ x^{3}-512=0 \] Choose the correct solution set below. A. $\left\{4\left(\cos 60^{\circ}+i \sin 60^{\circ}\right), 4\left(\cos 180^{\circ}+i \sin 180^{\circ}\right), 4\left(\cos 300^{\circ}+i \sin 300^{\circ}\right)\right\}$ B. $\left\{8\left(\cos 0^{\circ}+i \sin 0^{\circ}\right), 8\left(\cos 120^{\circ}+i \sin 120^{\circ}\right), 8\left(\cos 240^{\circ}+i \sin 240^{\circ}\right)\right\}$ C. $\left\{4\left(\cos 0^{\circ}+i \sin 0^{\circ}\right), 4\left(\cos 120^{\circ}+i \sin 120^{\circ}\right), 4\left(\cos 240^{\circ}+i \sin 240^{\circ}\right)\right\}$ D. $\left\{8\left(\cos 60^{\circ}+i \sin 60^{\circ}\right), 8\left(\cos 180^{\circ}+i \sin 180^{\circ}\right), 8\left(\cos 300^{\circ}+i \sin 300^{\circ}\right)\right\}$

Solution

Solution Steps

Step 1: Identify the Equation

We are given the equation:

\[ x^3 - 512 = 0 \]

This can be rewritten as:

\[ x^3 = 512 \]

Step 2: Find the Principal Root

The principal cube root of 512 is:

\[ x = \sqrt[3]{512} = 8 \]

Step 3: Use De Moivre's Theorem

To find all complex roots, we express 512 in polar form. Since 512 is a real number, it can be written as:

\[ 512 = 8^3 = 8(\cos 0^\circ + i \sin 0^\circ) \]

Using De Moivre's Theorem, the cube roots are given by:

\[ x_k = 8 \left( \cos \left( \frac{0^\circ + 360^\circ k}{3} \right) + i \sin \left( \frac{0^\circ + 360^\circ k}{3} \right) \right) \]

for $k = 0, 1, 2$.

Step 4: Calculate the Roots

For $k = 0$:

\[ x_0 = 8 \left( \cos 0^\circ + i \sin 0^\circ \right) = 8 \]

For $k = 1$:

\[ x_1 = 8 \left( \cos 120^\circ + i \sin 120^\circ \right) \]

For $k = 2$:

\[ x_2 = 8 \left( \cos 240^\circ + i \sin 240^\circ \right) \]

Final Answer

The solution set is:

\[ \boxed{\left\{8\left(\cos 0^{\circ}+i \sin 0^{\circ}\right), 8\left(\cos 120^{\circ}+i \sin 120^{\circ}\right), 8\left(\cos 240^{\circ}+i \sin 240^{\circ}\right)\right\}} \]

Thus, the correct choice is B.

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