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Question

If α,β are roots of ax2+bx+c=0 then 1α3+1β3=?

A
3abcb3a3
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B
3abb3a2c
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C
3abcb3c3
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D
b22acac
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Solution

The correct option is C 3abcb3c3
α and β are roots of the equation ax2+bx+c=0
αβ=ca ------ ( 1 )
α3β3=c3a3 ----- ( 2 )
α+β=ba ------ ( 3 )
(α+β)3=α3+β3+3αβ(α+β)
(ba)3=α3+β3+3(ca)(ba) [ By using ( 1 ) and ( 3 ) ]

b3a3=α3+β33bca2

α3+β3=b3a3+3bca2

α3+β3=b3+3abca3

α3+β3=3abcb3a3 ----- ( 4 )
Now,
1α3+1β3=α3+β3α3β3

=3abcb3a3c3a3 [ By using ( 2 ) and ( 4 ) ]

=3abcb3a3×a3c3

=3abcb3c3

1α3+1β3==3abcb3c3

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