Question

If alpha and beta are zeroes of quadratic polynomial f(x)= ax²+bx+c, evaluate 1/alpha³ + 1/beta³​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \: \: \bull\sf \: \: \: \alpha, \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \: \: \: \: \bull \sf \: \: \dfrac{1}{ { \alpha }^{3} } + \dfrac{1}{ { \beta }^{3} }$

$\large\underline{\sf{Solution-}}$

Given that

$\:\sf \: \: \: \alpha, \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c$

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha + \beta = - \: \dfrac{b}{a}$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha \beta = \dfrac{c}{a}$

Now,

Consider

$\rm :\longmapsto\: \sf \: \: \dfrac{1}{ { \alpha }^{3} } + \dfrac{1}{ { \beta }^{3} }$

$\sf \: \: = \: \dfrac{ { \alpha }^{3} + { \beta }^{3} }{ { \beta }^{3} { \alpha }^{3} }$

$\sf \: \: = \: \dfrac{ {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta )}{ {( \alpha \beta )}^{3} }$

$\sf \: \: = \: \dfrac{{\bigg( - \dfrac{b}{a} \bigg) }^{3} - 3{\bigg( \dfrac{c}{a} \bigg) }{\bigg( - \dfrac{b}{a} \bigg) }}{ {\bigg( \dfrac{c}{a} \bigg) }^{3} }$

$\sf \: \: = \: \dfrac{{\bigg( - \dfrac{ {b}^{3} }{ {a}^{3}} + \dfrac{3bc}{ {a}^{2} } \bigg) }}{\dfrac{ {c}^{3} }{ {a}^{3} } }$

$\sf \: \: = \: {\bigg( \dfrac{ - {b}^{3} + 3abc }{ \cancel{{a}^{3} }} \bigg) } \times \dfrac{ \cancel{ {a}^{3}} }{ {c}^{3} }$

$\sf \: \: = \: \dfrac{ { - b}^{3} + 3abc }{ {c}^{3} }$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \quad\bf \: \dfrac{1}{ { \alpha }^{3} } + \dfrac{1}{ { \beta }^{3} } = \dfrac{ - {b}^{3} + 3abc}{ {c}^{3} } \quad}}$

Additional Information :-

$\: \sf \:\alpha,\beta,\gamma\: are \: zeroes \: of \: f(x) = {ax}^{3} + {bx}^{2} + cx + d \: then$

$\blue{ \boxed{ \sf \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\blue{ \boxed{ \sf \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\blue{ \boxed{ \sf \: \alpha \beta \gamma = - \dfrac{d}{a}}}$