Question

`omega` is a cube root of unity. Show the following?

`(a+b+c)(a+bomega+comega^2)(a+bomega^2+comega) = a^3 + b^3 + c^3 - 3abc`

Answer 1

We are given that `omega` is a cube root of unity; therefore `omega` satisfies the equation

`z^3 = 1`

`:. z^3 - 1 = 0`
`:. (z-1)(z^2+z+1) = 0`

From which can get the actual values of the roots:

`z =1,(-1+-sqrt(3)i)/2`

Thus some easily verifiable properties are that:

• The roots can be denoted by `1, omega, omega^2` where `omega` is any of the complex roots.
• `1+omega+omega^2=0 => omega+omega^2=-1`.
• `omega^3=1 => omega^4=omega`. etc

So now, let us consider the RHS of the given expression:

`RHS = (a+b+c)(a+bomega+comega^2)(a+bomega^2+comega)`

`" " = (a+b+c) (a^2+abomega^2+acomega+abomega+b^2omega^3+bcomega^2+ acomega^2+bcomega^4+c^2omega^3)`

`" " = (a+b+c) (a^2+b^2+c^2+acomega+abomega+bcomega+abomega^2+bcomega^2+ acomega^2+`

`" " = (a+b+c) (a^2+b^2+c^2+ac(omega+omega^2) + ab(omega+omega^2) + bc(omega+omega^2) )`

`" " = (a+b+c) (a^2+b^2+c^2 - ac - ab - bc)`

`" " = a^3+ab^2+ac^2 - a^2c - a^2b - abc+a^2b+b^3+bc^2 - abc - ab^2 - b^2c+a^2c+b^2c+c^3 - ac^2 - abc - bc^2`

And the almost all cross terms cancel, leaving:

`RHS = a^3 + b^3 + c^3 - 3abc \ \ \` QED