`omega` is an imaginary root of unity.
Prove that
(i) `( a + bomega + comega^(2))^(3) + (a+bomega^(2) + comega)^(3) = (2a-b-c)(2b -a -c)(2c -a-b)`
(ii) If `a+b+c = 0` then prove that `(a + bomega + comega^(2))^(3)+(a+bomega^(2) + comega)^(3) = 27abc`.

Explanation

(i) Let `a + bomega + comega^(2) = x` and `a + bomega^(2) + comega = y` , Then
`(a+bomega + comega^(2))^(3) + (a+ bomega^(2) + comega)^(3) = x^(3) + y^(3)`
`=(x+y)(x+ omegay)(x+ omega^(2)y)`
Now,
`x + y = (a+ bomeag + comega^(2) )+ (a+ bomega^(2) + comega)`
`=2a + b(omega+omega^(2))`
`=2-b-c`
`x + omegay= (a+ bomega+ comega^(2)) + omega(a+bomega^(2) + comega)`
`= (1+omega) a+ 1(1+ omega)b + 2omega^(2)c`
`omega^(2) (2c-a-b)`
Similarly,
`x + omega^(2)y = omega(2b-a-c)`
`rArr (x+y)(x+ omegay)(x+ omega^(2)y)`
`=omega^(3)(2a-b-c)(2c-a-b)(2b -a-c)`
`=(2a-b-c)(2c-a-b)(2b-a-c)`
(ii) `a+ b+c=0`
. `rArr b+c = -a,c+a= -b` and `a+ b= -c`
Putting these values of the R.H.S of result (i), we get
`(a+ bomega + comega^(2))^(3) + (a+bomega^(2)+ comega)^(3) = 27abc`