Answer :
To evaluate the cube root of [tex]\(-27\)[/tex], we need to determine what number, when raised to the power of 3, equals [tex]\(-27\)[/tex]. Here, we are looking for the complex cube root because [tex]\( \sqrt[3]{-27} \)[/tex] can lead to complex solutions.
1. Understanding the problem:
- We are given [tex]\( \sqrt[3]{-27} \)[/tex].
- We need to find a number [tex]\( x \)[/tex] such that [tex]\( x^3 = -27 \)[/tex].
2. Formulating in polar form:
- The complex number [tex]\(-27\)[/tex] can be represented in polar form. [tex]\(-27\)[/tex] on the complex plane is in quadrant II, which can be represented as [tex]\( 27 \text{cis}(\pi) \)[/tex], where [tex]\(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\)[/tex].
3. Finding the principal root:
- For complex numbers, the cube root is obtained by taking the cube root of the magnitude and dividing the angle by 3.
- The magnitude (absolute value) of [tex]\(-27\)[/tex] is 27.
- The argument (angle) is [tex]\(\pi\)[/tex].
- The principal cube root of 27 is [tex]\( 27^{1/3} = 3 \)[/tex].
- The cube root of the angle is [tex]\( \frac{\pi}{3} = \frac{\pi}{3} \)[/tex].
4. Including the principal cube root (r, θ/3):
- Converting back to the rectangular form (x + yi), we have:
[tex]\[ 3 \text{cis}\left(\frac{\pi}{3}\right) = 3 \left( \cos\left( \frac{\pi}{3} \right) + i \sin\left( \frac{\pi}{3} \right) \right) \][/tex]
5. Calculating the trigonometric functions:
- [tex]\( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \)[/tex]
- [tex]\( \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \)[/tex]
- Plugging these values in:
[tex]\[ 3 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{3}{2} + i \frac{3\sqrt{3}}{2} \][/tex]
6. Checking the complex cube roots:
- For the given value [tex]\(-27\)[/tex], there are actually three cube roots, but we only need the principal one.
- Using these steps, one of the cube roots is:
[tex]\[ x = \frac{3}{2} + i \frac{3\sqrt{3}}{2} \][/tex]
- Rewriting this as a single term:
[tex]\[ x = 1.5000000000000004 + 2.598076211353316i \][/tex]
Therefore, the cube root of [tex]\( -27 \)[/tex] is [tex]\( \left(1.5000000000000004+2.598076211353316i\right) \)[/tex].
1. Understanding the problem:
- We are given [tex]\( \sqrt[3]{-27} \)[/tex].
- We need to find a number [tex]\( x \)[/tex] such that [tex]\( x^3 = -27 \)[/tex].
2. Formulating in polar form:
- The complex number [tex]\(-27\)[/tex] can be represented in polar form. [tex]\(-27\)[/tex] on the complex plane is in quadrant II, which can be represented as [tex]\( 27 \text{cis}(\pi) \)[/tex], where [tex]\(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\)[/tex].
3. Finding the principal root:
- For complex numbers, the cube root is obtained by taking the cube root of the magnitude and dividing the angle by 3.
- The magnitude (absolute value) of [tex]\(-27\)[/tex] is 27.
- The argument (angle) is [tex]\(\pi\)[/tex].
- The principal cube root of 27 is [tex]\( 27^{1/3} = 3 \)[/tex].
- The cube root of the angle is [tex]\( \frac{\pi}{3} = \frac{\pi}{3} \)[/tex].
4. Including the principal cube root (r, θ/3):
- Converting back to the rectangular form (x + yi), we have:
[tex]\[ 3 \text{cis}\left(\frac{\pi}{3}\right) = 3 \left( \cos\left( \frac{\pi}{3} \right) + i \sin\left( \frac{\pi}{3} \right) \right) \][/tex]
5. Calculating the trigonometric functions:
- [tex]\( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \)[/tex]
- [tex]\( \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \)[/tex]
- Plugging these values in:
[tex]\[ 3 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{3}{2} + i \frac{3\sqrt{3}}{2} \][/tex]
6. Checking the complex cube roots:
- For the given value [tex]\(-27\)[/tex], there are actually three cube roots, but we only need the principal one.
- Using these steps, one of the cube roots is:
[tex]\[ x = \frac{3}{2} + i \frac{3\sqrt{3}}{2} \][/tex]
- Rewriting this as a single term:
[tex]\[ x = 1.5000000000000004 + 2.598076211353316i \][/tex]
Therefore, the cube root of [tex]\( -27 \)[/tex] is [tex]\( \left(1.5000000000000004+2.598076211353316i\right) \)[/tex].
