Year 12s: what will be the first way you research your future options? >>

FP2 question

Exercise 3H, question 2(iv)

2. If w is a complex cube root of unity, w ≠ 1, prove that:

(a+b+c)(a+wb+(w^2)c)(a+(w^2)b+wc) = a^3 + b^3 + c^3 - 3abc

Is the only way of doing this question multiplying out all the brackets and then using roots of unity to cancel? Because I feel like there must be a faster way but I can't figure out what it is.

Reply 1

ghostwalker

Original post by klosovic

Because I feel like there must be a faster way but I can't figure out what it is.

Inclined to agree, but I can't see it.

Reply 2

davros

Original post by klosovic

It's not too heavy to do directly tbh.

You can 'see' that the a^3 term is what it is; similarly you get b(wb)(w^2b) which gives you w^3b^3 which you know is b^3, and similarly for the c^3 term.

Then look at how you can get terms involving a^2b or a^2c and what happens to the coefficients of these.

The only term with lots of contributions is the 'abc' term, and you can still write down the separate terms that go into this fairly quickly.