Question

ans is 5 but I wanna know the method . U can also briefly explain, thanks​

Answer 1

Given -

$\alpha, \: \beta, \: \gamma \: are \: the \: zeroes \: of \: p(x) = {6x}^{3} + {3x}^{2} - 5x + 1 = 0$

To find -

$the \: value \: of \: ( \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } )$

Solution -

${6x}^{3} + {3x}^{2} - 5x + 1 = 0$

Here,

a=6

b=3

c= -5

d=1

$\alpha, \: \beta, \: and \: \gamma \: are \: zeroes \: of \: given \: polynomial$

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 3}{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1}{2}$

$\alpha \beta \gamma = \frac{ - d}{a} = \frac{ - 1}{6}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = \frac{ - 5}{6}$

Now,

$= ( \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } )$

$= \frac{ \beta \gamma + \alpha \gamma + \beta \gamma }{ \alpha \beta \gamma }$

$= \frac{( \frac{c}{a}) }{( \frac{ - d}{a}) }$

$= \frac{ (\frac{ - 5}{6}) }{( \frac{ - 1}{6}) }$

$= \frac{ - 5}{6} \div \frac{ - 1}{6}$

$= \frac{ - 5}{6} \times - 6$

$= 5$

$\therefore \: the \: value \: of \: ( \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } ) \: is \: 5$