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Prove `: 2(sin^(6)theta+cos^(6)theta)-3(sin^(4)theta+cos^(4)theta)+1=0`.

Answer» Proof: `LHS=2(sin^(6)theta+cos^(6)theta)-3(sin^(4)theta)+cos^(4)theta)+1`
`=2[(sin^(2)theta)^(3)+(cos^(2)theta)^(3)]-3(sin^(4) theta+cos^(4) theta)+1`
`=2[(sin^(2)theta+cos^(2) theta)(sin^(4) theta-sin^(2) theta.cos^(2) theta+cos^(4) theta)]-3(sin^(4)theta+cos^(4)theta)+1`
`......[(a^(3)+b^(3))=(a+b)(a^(2)-ab+b^(2)]`
`=2(1)(sin^(4) theta-sin^(2)theta.cos^(2) theta+cos^(4)theta)-3(sin^(4) theta+cos^(4)theta)+1`
`=2sin^(4) theta-2sin^(2) theta. cos^(2) theta+2cos^(4) theta-3sin^(4) theta-3cos^(4) theta+1`
`=-sin^(4) theta-cos^(4) theta-2sin^(2) theta. cos^(2) +1`
`=-(sin^(4) theta+2sin^(2) theta.cos^(2) theta+cos^(4) theta)+1`
`=-(sin^(2) theta+cos^(2) theta)^(2) +1`...........`[(a^(2) +2ab+b^(2))=(a+b)^(2)]`
`=-(1)^(2)=1`..........`(sin^(2) theta+cos^(2) theta=1)`
`=-1+1=0=RHS`
`:.2(sin^(6)theta+cos^(6)theta)-3(sin^(4) theta+cos^(4) theta)+1=0`.


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