1.

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»


Solution :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)`


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