if alpha,beta are the roots of x^2+2x-1=0.find the values of alpha cubed + beta cubed

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Solution

$G i v e n q u a d r a t i c e q u a t i o n i s x^{2} + 2 x - 1 = 0 N o w i t i s g i v e n t h a t α a n d β a r e t h e r o o t s o f t h i s e q u a t i o n . N o w a s w e k n o w t h a t s u m o f r o o t s o f a q u a d r a t i c e q u a t i o n i s - \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f x^{2}} ∴ α + β = - 2 . . . . . . . . . . (1) A l s o w e k n o w t h a t p r o d u c t o f r o o t s o f a q u a d r a t i c e q u a t i o n i s \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f x^{0} (c o n s \tan t t e r m)} ∴ α β = - 1 . . . . . . . (2) N o w α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β) N o w p u t t i n g t h e v a l u e s o f α + β a n d α β, w e g e t α^{3} + β^{3} = (- 2)^{3} - 3 \times (- 1) \times (- 2) = - 8 - 6 = - 14$

Alternate method:-

$x^{2} + 2 x - 1 = 0 \Rightarrow x = \frac{- 2 \pm \sqrt{(2)^{2} - 4 \times 1 \times (- 1)}}{2} = \frac{- 2 \pm \sqrt{8}}{2} = \frac{- 2 \pm 2 \sqrt{2}}{2} ∴ α = - 1 + \sqrt{2} a n d β = - 1 - \sqrt{2} N o w α^{3} + β^{3} = (- 1 + \sqrt{2})^{3} + (- 1 - \sqrt{2})^{3} = - 14$

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