If `a= .^(20)C_(0) + .^(20)C_(3) + .^(20)C_(6) + .^(20)C_(9) + "…..", ``b = .^(20)C_(1) + .^(20)C_(4) + .^(20)C_(7) + "……"' and ``c = .^(20)C_(2) + .^(20)C_(5) + .^(20)C_(8) + "…..",` then
Value of `a^(3) + b^(3) + c^(3) - 3abc` is

Answer

B

Explanation

We have
`a+b+c = 2^(30)`
Now, `a^(3) + b^(3) + c^(3) - c^(3) - 3abc`
`= (a+b+c) (a+bomega + comega^(2)) (a +bomega^(2) + comega)`
`= 2^(20) (1+omega)^(20) (1+omega^(2))^(20)`
`= 2^(20)`