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If a^3+b^3+6a b c=8c^3&omega is a cube r...

If `a^3+b^3+6a b c=8c^3&omega` is a cube root of unity then: `a , b , c` are in `AdotPdot` (b) `a , b , c ,` are in `HdotPdot` `a+bomega-2comega^2=0` `a+bomega^2-2comega=0`

A

`a,c,b` are in A.P

B

a,c,b are in H.P

C

`a+bomega - 2comega^(2) = 0`

D

`a+ bomega^(2) -2comega = 0`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
We have `a^(3) + b^(3) - 8c^(3) + 6abc = 0`
`rArr a^(3) + b^(3) + (-2c)^(3) - 3ab(-2c) = 0`
`rArr (a + b- 2c) (a+bomega - 2comega^(2)) (a+ boemga^(2)-2comega) =0`
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