Question

Find all thecube roots of 8​

Answer 1

Answer:

Step-by-step explanation:

Cuberoot of 8

2*2*2=2

Answer 2

Since you're mentioning "all" the cube roots, I'm assuming you're asking about complex roots as well.

In that case, it is more convenient to use De Moivre's form:

$z=8=8(1+0i)=8[\cos(2\pi)+i\sin(2\pi)]$

Which implies

$\sqrt[3]{z}=2\left[\cos\left(\dfrac{2n\pi}{3}\right)+i\sin\left(\dfrac{2n\pi}{3}\right)\right],\quad n=0, 1, 2$

Which yields the roots

$z_0 = 2\left[\cos\left(0\right)+i\sin\left(0\right)\right] = 2$

$z_1=2\left[\cos\left(\dfrac{2\pi}{3}\right)+i\sin\left(\dfrac{2\pi}{3}\right)\right]=-1+i\sqrt{3}$

$z_2=2\left[\cos\left(\dfrac{4\pi}{3}\right)+i\sin\left(\dfrac{4\pi}{3}\right)\right]=-1-i\sqrt{3}$