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

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SUMMARY

The discussion centers on proving that the expression x + wy + w2z is a factor of the polynomial x3 + y3 + z3 - 3xyz, where w represents a complex cube root of 1, specifically w = -1/2 ± (√3/2)i. Participants suggest using polynomial division to find the remainder when dividing the polynomial by the factor, and they reference the relationship z3 - 1 = (z - 1)(z2 + z + 1) to derive further insights about the roots of the polynomial.

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Ferrus
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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.
 
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Hi Ferrus, welcome to PF:smile:

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

Hint 2: If [itex]z^3=1[/itex], then [itex]z^3-1=(z-1)(z^2+z+1)=0[/itex]... so what can you say about [itex]w^2+w+1[/itex]?
 
Welcome to PF!

Ferrus said:
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! :smile:

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

so put x = … ? :wink:
 

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