root(6)(-64)=?Please,give all the possible answers.

May 23, 2018

See beow

Explanation:

Calculate $\sqrt[6]{- 64}$ means you have to find a real number $x$ such that ${x}^{6} = - 64$. Such number doesnt exist because if it were positive, then never will get a negative number as product, if it were negative, then

(-x)·(-x)·(-x)·(-x)·(-x)·(-x)= positive number (there are an even number of factors (6) and never will get $- 64$)

In summary that $\sqrt[6]{- 64}$ has no real solutions. There is no number $x$ such that ${x}^{6} = - 64$

But in complex set of numbers there are 6 solutions

First put $- 64$ in polar form which is ${64}_{180}$

Then the six solutions ${r}_{i}$ from i=0 to i=5 are

${r}_{0} = {\sqrt[6]{64}}_{\frac{180}{6}} = {2}_{30}$
${r}_{1} = {\sqrt[6]{64}}_{\frac{180 + 360}{6}} = {2}_{90}$
${r}_{2} = {2}_{\frac{180 + 720}{6}} = {2}_{150}$
${r}_{3} = {2}_{\frac{180 + 1080}{6}} = {2}_{210}$
${r}_{4} = {2}_{270}$
${r}_{5} = {2}_{330}$

Who are these numbers?

${r}_{0} = 2 \left(\cos 30 + i \sin 30\right) = \sqrt{3} + i$
${r}_{1} = 2 i$
${r}_{2} = - \sqrt{3} + i$
${r}_{3} = - \sqrt{3} - i$
${r}_{4} = - 2 i$
${r}_{5} = \sqrt{3} - i$