If α,β,γ are the roots of x3 + px 2 + qx + r = 0, then α−1βγβ−1γαγ−1αβ=

Here,α+β+γ=−p,αβ+αγ+βγ=q,αβγ=−rα−1βγβ−1γαγ−1αβ =α−ααβγβ−βαβγγ−γαβγ =αβγ1−1αβγ3 =−r1+1r3 =−(1+r)3r2

If α,β,γ are the roots of x3 + px 2 + qx + r = 0, then α−1βγβ−1γαγ−1αβ=

Here,α+β+γ=−p,αβ+αγ+βγ=q,αβγ=−rα−1βγβ−1γαγ−1αβ =α−ααβγβ−βαβγγ−γαβγ =αβγ1−1αβγ3 =−r1+1r3 =−(1+r)3r2

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If $α, β, γ$ are the roots of x³ + px ² + qx + r = 0, then $(α - \frac{1}{β γ}) (β - \frac{1}{γ α}) (γ - \frac{1}{α β}) =$

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$\frac{(1 + r)^{3}}{r^{2}}$

$- \frac{(1 + r)^{3}}{r}$

$- \frac{(1 + r)^{3}}{r^{2}}$

$\frac{(1 + r)^{3}}{r}$

answer is C.

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Detailed Solution

Here,

$\begin{array}{l} α + β + γ = - p, α β + α γ + β γ = q, α β γ = - r \\ (α - \frac{1}{β γ}) (β - \frac{1}{γ α}) (γ - \frac{1}{α β}) \\ = (α - \frac{α}{α β γ}) (β - \frac{β}{α β γ}) (γ - \frac{γ}{α β γ}) \\ = α β γ {(1 - \frac{1}{α β γ})}^{3} \\ = - r {(1 + \frac{1}{r})}^{3} \\ = - \frac{(1 + r)^{3}}{r^{2}} \end{array}$