Question

solve the equation. x^3 - 1=0. i know it means find the third roots of 1.

Answer 1

the set up is supposed to look like this for ex: [32(cos(2pi) + isin(2pi)]^1/5

Answer 2

you factor the left side using the difference of cubes rule

x^3 - 1=0

x^3 - 1^3 =0

(x-1)(x^2 + x*1 + 1^2) = 0

(x - 1)(x^2 + x + 1) = 0

Answer 3

then you use the zero product property to get to

x-1 = 0 or x^2 + x + 1 = 0

Answer 4

we didn't learn it that way for this type of problem. its power and roots of complex numbers

Answer 5

another method

x^3 - 1 = 0

x^3 = 1

x^3 = 1 + 0i

1 + 0i is in rectangular or cartesian form

what is 1+0i in trignometric form?

Answer 6

wouldn't the answer be 1 then?

Answer 7

in other words, are you able to convert 1+0i to the form r*[ cos(theta) + sin(theta) ] ?

Answer 8

i think so cuz i did that

Answer 9

what did you get

Answer 10

3

Answer 11

1+0i is at the point (1,0)

Answer 12

how far is that point away from the origin?
what is the angle?

Answer 13

1 point away n 0 is the angle

Answer 14

is theta 0

Answer 15

so r = 1 and theta = 0

Answer 16

i get that

Answer 17

1+0i = 1*[cos(0) + sin(0)]

Answer 18

you'll then use de moivre's theorem

Answer 19

but the root is 3 so doesn't root three go at the end?

Answer 20

oh wait nevermind, no you dont

Answer 21

please you ahve to use complex numbers, so please write 1 as below:

\[1 = \cos \left( {2\pi + 2k\pi } \right) + i\sin \left( {2\pi + 2k\pi } \right)\]

Answer 22

oops... you have...

Answer 23

hint:
x^3-1=0, becomes:
\[x=\sqrt[3]{1}\], or:
\[\begin{gathered}
x = \cos \left( {\frac{{2\pi + 2k\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi + 2k\pi }}{3}} \right) \hfill \\
k = 0,1,2 \hfill \\
\end{gathered} \]

Answer 24

now, please substitute k=0,1,2 into the above formula and you will get your three roots, each for each value of k

Answer 25

ok thnx

Answer 26

thanks!