Question

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

Answer 1

See beow

Explanation:

Calculate `root(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 `root(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=root(6)64_(180/6)=2_30`
`r_1=root(6)64_((180+360)/6)=2_90`
`r_2=2_((180+720)/6)=2_150`
`r_3=2_((180+1080)/6)=2_210`
`r_4=2_270`
`r_5=2_330`

Who are these numbers?

`r_0=2(cos30+isin30)=sqrt3+i`
`r_1=2i`
`r_2=-sqrt3+i`
`r_3=-sqrt3-i`
`r_4=-2i`
`r_5=sqrt3-i`