Question

if alpha beta are the zeros of x square - 4 x + 1 then one by Alpha cube plus one by beta cube equals​

Answer 1

$\boxed{\red{\frak{question}}}$

$\tt \: if \: \alpha and \: \beta \: are \: zeroes \: of \: \\ \tt \: {x}^{2} - 4x + 1 \: then \: find \\ \tt \: \frac{1}{ { \alpha }^{3} } + \frac{1}{ { \beta }^{3} } .$

$\boxed{\red{\frak{solution}}}$

$\tt \: comparing \: {x}^{2} - 4x + 1 \: with \: \\ \tt \: a {x}^{2} + bx + c \: \\ \tt \: we \: will \: get \\ \\ \tt \: a = 1 \\ \tt \: b = - 4 \\ \tt \: c = 1$

As we know,

$\tt \: sum \: of \: zeroes \: = \frac{ - b}{a} \\ \\ \tt \: \alpha + \beta = \frac{ - ( - 4)}{1} = 4 \: \: - - eqn(1)$

also,

$\tt \: product \: of \: zeroes \: = \frac{c}{a} \\ \\ \tt \: \alpha \beta = \frac{1}{1} = 1 \: \: - - - eqn(2)$

$\tt \: we \: have \: to \: find \\ \\ \tt \: \frac{1}{ { \alpha }^{3} } + \frac{1}{ { \beta }^{3} } \\ \\ \tt \: taking \: lcm \\ \\ \frac{ { \alpha }^{3} + { \beta }^{3} }{ { \alpha }^{3} { \beta }^{3} } \\ \\ \tt \: using \: identity \: \\ \tt \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ \\ \tt \: = \frac{ { (\alpha + \beta) }^{3} - 3 \alpha \beta ( \alpha + \beta ) }{( { \alpha \beta) }^{3} } \\ \\ \tt \: using \: eqn(1) \: and \: eqn(2) \\ \\ \tt \: = \frac{ {4}^{3} - 3(1)(4) }{ {1}^{3} } \\ \\ \tt \: = \frac{64 - 12}{1} = 52$