Question

If α and β are zeroes of quadratic polynomial. f(x) = x² -x -4 i ) α³ + β³ ii ) 1/α³ + 1/β³

Answer 1

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Answer 2

$\tt since \: \alpha \: and \: \beta \: are \: zeroes \: of \: quadratic \: polynomial \: \: .$

$\tt \: f(x) = {x}^{2} - x - 4$

$\tt \therefore \: \: \: \alpha + \beta = ( \frac{ - b}{a} ) = 1 \: and \: \alpha \: \beta = \frac{c}{a} = \frac{ - 4}{1} = - 4$

$\tt i) \alpha ^{3} + \beta {}^{3} = {( \alpha + \beta })^{3 } - 3 \alpha \beta ( \alpha + \beta )$

$\alpha {}^{3} + \beta {}^{3} = ( \frac{ - b}{a} ) - 3 \frac{c}{a} ( - \frac{b}{a} )$

$\longrightarrow \: \alpha {}^{3} + \beta {}^{3} = - (1 {)}^{3} - 3( - 4)(1) = 13$

$\frac{1}{ {a}^{3} } + \frac{1}{ {b +}^{3} } = \frac{ { \alpha {}^{3} + { \beta }^{3} } }{ { \alpha }^{3} + \beta {}^{3} } = \frac{ \frac{3abc - {b}^{3} }{ {a}^{3} } }{ \frac{( \frac{c}{a {}^{3} } )}{} }$

$\rightarrow \: \frac{1}{ { \alpha }^{3} } + \frac{1}{ { \beta }^{3} } = \frac{13}{( - 4) {}^{3} } = \frac{ - 13}{64}$