# if alpha beta and gamma are the zeroes of polynomial 6x3+3x2-5x+1 then find the value of alpha raised to -1+beta raised to -1+gamma raised to -1

We have $\alpha$ , $\beta$  And $\gamma$ are the zeros of cubic polynomial 6x3 + 3x2 - 5x + 1
We know by the relationship between zeros and coefficient of a cubic polynomial , is
Sum of zeros =

$\alpha$ + $\beta$  +  $\gamma$   = $-\frac{3}{6}$

$\alpha$ + $\beta$  +  $\gamma$  = $-\frac{1}{2}$                                      -------------------  ( 1 )

Sum of the products of zeros taken two at a time =
$\alpha$ $\beta$  +  $\beta$ ​$\gamma$  + $\gamma$ $\alpha$   = $\frac{-5}{6}$             ---------------------  ( 2 )

Product of zeros  =
$\alpha$ $\beta$ $\gamma$   = $-\frac{1}{6}$                                   --------------------- ( 3 )

Now for value of

$\alpha$-1  + $\beta$-1  + $\gamma$-1

We can simplify it As :

$⇒$
Taking L.C.M. and get

$⇒$

Now substitute values from equation 2 and 3 we get

$⇒$$\frac{\frac{-5}{6}}{-\frac{1}{6}}$

$⇒$  $\alpha$-1 + $\beta$-1 + $\gamma$-1    =  5                                           ( Ans )

• 185

The given polynomial is

6x3 + 3x2 - 5x + 1

Here a = 6, b = 3, c = -5 and d = 1

α + β + γ = -b/a = -3/6 = -1/2

αβγ = -d/a = -1/6

α-1 + β-1 + γ-1 = 1/α + 1/β + 1/γ

= (α + β + γ)/ αβγ

= (-1/2)/(-1/6)

= 3

• -22
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