If x 3-1 can be written as x -0 x -1 x -2 - Where- -0- -1 and -2 are the roots of the equation and 1-3-0-1-3-1-1-3-2- k -26- Find the value of k-

Given (x^3 1) = (x α _0) (x α _1) (x α _2)..........(1) We see that we have relation in 1/(x α _0) + 1/(x α _1) + 1/(x α _2) How to get that rel ...

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Standard XII

Mathematics

Sequence

If x 3-1 can ...

Question

If $x^{3}$ - 1 can be written as $(x - α_{0})$ $(x - α_{1})$ $(x - α_{2})$ .Where, $α_{0}$ , $α_{1}$ and $α_{2}$ are the roots of the equation

and $\frac{1}{(3 - α_{0})}$ + $\frac{1}{(3 - α_{1})}$ + $\frac{1}{(3 - α_{2})}$ = $\frac{k}{26}$ . Find the value of k.

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Solution

Given ( $x^{3}$ - 1) = $(x - α_{0})$ $(x - α_{1})$ $(x - α_{2})$ ..........(1)

We see that we have relation in $\frac{1}{(x - α_{0})}$ + $\frac{1}{(x - α_{1})}$ + $\frac{1}{(x - α_{2})}$

How to get that relation?

If we take the log on equation 1 and then differentiate it we will get the relation $\frac{1}{(x - α_{0})}$ + $\frac{1}{(x - α_{1})}$ + $\frac{1}{(x - α_{2})}$

Taking log on both sides

$log (x^{3} - 1)$ = $log (x - α_{0})$ + $log (x - α_{1})$ + $log (x - α_{2})$

Differentiating on both sides

$\frac{1}{x^{3} - 1}$ . 3 $x^{2}$ = $\frac{1}{(x - α_{0})}$ + $\frac{1}{(x - α_{1})}$ + $\frac{1}{(x - α_{2})}$

Put n = 3

$\frac{3. 3^{2}}{27 - 1}$ = $\frac{1}{(3 - α_{0})}$ + $\frac{1}{(3 - α_{1})}$ + $\frac{1}{(3 - α_{2})}$ = $\frac{27}{(26)}$

k = 27

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Similar questions

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0 \leq k \leq n - 1

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and if the equation having

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If x3 - 1 can be written as (x−α0) (x−α1) (x−α2).Where,α0,α1 and α2 are the roots of the equation and 1(3−α0) + 1(3−α1) + 1(3−α2) = k26. Find the value of k. __

If $x^{3}$ - 1 can be written as $(x - α_{0})$ $(x - α_{1})$ $(x - α_{2})$ .Where, $α_{0}$ , $α_{1}$ and $α_{2}$ are the roots of the equation

and $\frac{1}{(3 - α_{0})}$ + $\frac{1}{(3 - α_{1})}$ + $\frac{1}{(3 - α_{2})}$ = $\frac{k}{26}$ . Find the value of k.

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