Question

By solving the equation x^3+1=0, find the three cube roots of -1. If one of the complex cube roots is A express in terms of A. Prove that 1+A^2=A.

Answer 1

Without using complex roots?

Answer 2

using complex roots

Answer 3

Using this
\[\Large a+bi=r(\cos\theta \ + \ i\sin\theta)\]form?

Answer 4

why bother

Answer 5

Yeah, why bother :D
We can factor x^3 + 1
as it's the sum of two cubes :)

Answer 6

\[x^3+1=(x+1)(x^2-x+1)\] quadratic formula gives the second two zeros

Answer 7

could you be more specific?

Answer 8

\[\Large x^3+1=(x+1)(x^2-x+1)\]
Which means -1 is one of the roots of the equation right?

Answer 9

So the other two roots are whatever the roots of this equation is...
\[\Large x^2-x+1=0\]
Easily obtainable through the quadratic formula :)

Answer 10

ok, thx