Question

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Part 1
Find all the cube roots of minus−8
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The real cube​ root(s) of minus−8 ​is/are enter your response here.
​(Simplify your answer. Use a comma to separate answers as​ needed.)
B.
There are no real cube roots of minus−8

Answer 1

To find the cube roots of \(-8\), we can start with the equation:

\[
x^3 = -8
\]

The real cube root of \(-8\) can be found by solving:

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

Calculating the cube root, we see that:

\[
x = -2
\]

Thus, the real cube root of \(-8\) is \(-2\).

Now, we should consider the complex cube roots as well. The cube roots of a number can be expressed in polar form. In polar coordinates:

- The modulus (magnitude) of \(-8\) is \(8\).
- The argument (angle) of \(-8\) is \(\pi\) (since it is located on the negative x-axis).

The general formula for the \(n\)th roots of a complex number in polar form \(r(\cos \theta + i\sin \theta)\) is given by:

\[
\sqrt[n]{r} \left( \cos \left(\frac{\theta + 2k\pi}{n}\right) + i \sin \left(\frac{\theta + 2k\pi}{n}\right) \right), \quad k = 0, 1, \ldots, n-1
\]

For \(-8\) (which we write as \(8(\cos \pi + i \sin \pi)\)), the cube roots can be calculated with \(n = 3\):

1. Finding the modulus:

\[
\sqrt[3]{8} = 2
\]

2. Finding the three angles for \(k = 0, 1, 2\):

- For \(k = 0\):

\[
\theta_0 = \frac{\pi + 2(0)\pi}{3} = \frac{\pi}{3} \quad \Rightarrow \quad 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) = 2\left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3}
\]

- For \(k = 1\):

\[
\theta_1 = \frac{\pi + 2(1)\pi}{3} = \pi \quad \Rightarrow \quad 2\left(\cos \pi + i \sin \pi\right) = 2(-1 + 0) = -2
\]

- For \(k = 2\):

\[
\theta_2 = \frac{\pi + 2(2)\pi}{3} = \frac{5\pi}{3} \quad \Rightarrow \quad 2\left(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3}\right) = 2\left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = 1 - i\sqrt{3}
\]

Thus, the three cube roots of \(-8\) are:

1. \( -2 \) (the real cube root),
2. \( 1 + i\sqrt{3} \),
3. \( 1 - i\sqrt{3} \).

**Final answer:**

A. The real cube root of \(-8\) is \(-2\).

If asked for all cube roots as well, they would be \(-2\), \(1 + i\sqrt{3}\), and \(1 - i\sqrt{3}\).