miscellaneous equation

If α and β are the complex cube roots of unity, and x = a + b, y = aα+ bβ and z = aβ + bα, then the value of xyz =.............................

Answer:

x³ = 1, or (x³ – 1) = 0, or (x – 1)(x² + x + 1) = 0

Now, (x² + x + 1) = 0

Or, (x² + 2.x.½ + ¼ - ¼ + 1) = 0

Or, x = - ½ ± i√3/2

Therefore, y = aα+ bβ = a(i√3 – 1)/2 - b(i√3 + 1)/2

Also, z = aβ + bα = - a(i√3 + 1)/2 + b(i√3 – 1)/2

So, xyz = (a + b) [a(i√3 – 1) - b(i√3 + 1)] [- a(i√3 + 1) + b(i√3 – 1)]/4

Or, xyz = (a+ b) [4a² + ab.( -2 – 2i√3) + ab.(-2 + 2i√3) + 4b²]/4

Or, xyz = (a+ b) [4a² - 4ab + 4b²]/4

Or, xyz = 4.(a+ b) [a² - ab +b²]/4

Or, xyz = a³ + b³