Proving a Factor of Complex Cube Root of 1 in x^3 + y^3 + z^3 - 3xyz Equation

[Ferrus]

Homework Statement

If w is a complex cube root of 1, prove that x + wy + w^2z is a factor of x^3+ y^3 + z^3 - 3xyz, and hence factorise the equation completely.

Homework Equations

Complex cube root of 1 = -1/2 +/- 3^1/2/2 i

The Attempt at a Solution

Erm, I feel way over my head. I have tried plugging in the equation to the first one but this doesn't seem to generate anything intelligible for me.

 

[gabbagabbahey]

Hi Ferrus, welcome to PF

Hint 1: If x + wy + w^2z is a factor of x^3+ y^3 + z^3 - 3xyz, what can you say about the remainder of \frac{x^3+ y^3 + z^3 - 3xyz}{x + wy + w^2z}

Hint 2: If z^3=1, then z^3-1=(z-1)(z^2+z+1)=0... so what can you say about w^2+w+1?

 

[tiny-tim]

Welcome to PF!

If w is a complex cube root of 1, prove that x + wy + w^2z is a factor of x^3+ y^3 + z^3 - 3xyz …

Hi Ferrus ! Welcome to PF!

Hint: if (x+a) is a factor of a polynomial, then put x = -a and the polynomial will be zero …

so put x = … ?