If the sum of the squares of the reciprocals of the roots and of the equation 3 x 2 + x - 1 = 0 is 15, then 6 ( 3 + 3 ) 2 =

Question

If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 {(α^{3} + β^{3})}^{2} =$

Smart Achievers · Accepted Answer

(24)

$Here α + β roots of equation$

$3 x^{2} + λ x - 1 = 0$

$α + β = \frac{- λ}{3}, α β = \frac{- 1}{3}$

$\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{{(α + β)}^{2} - 2 α β}{α^{2} β^{2}} = 15$

$λ^{2} = 9$

$Now$

$6 {(α^{3} + β^{3})}^{2} = 6 {(α + β)}^{2} ({{(α + β)}^{2} - 3 α β)}^{2}$

$= 6 (\frac{λ^{2}}{9}) {\frac{λ^{2}}{9} + 1}^{2} = 24$

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