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Question

If αandβ are zeroes of 2x29x7, find the values of: α3+β3,(αβ)2

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Solution

2x29x7
α,β are zeroes
So, α+β=92,αβ=72
α3+β3=(α+β)33(α+β)(αβ)
=721839272
=7218+1894=729+3788
=11078
(αβ)2=(α+β)24αβ
=(92)24(72)=814+282
=1374

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