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Nature of Roots

lf α, β, γ ...

Question

lf $α, β, γ$ are the roots of $x^{3} - 7 x + 6 = 0$ , then the equation whose roots are $(α - β)^{2}, (β - γ)^{2}, (γ - α)^{2}$ is:

A

(x - 7)^{3} - 21 (x - 7)^{2} + 972 = 0

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B

(x + 1)^{3} - 21 (x + 1)^{2} + 972 = 0

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C

(x + 7)^{3} - 21 (x + 7)^{2} + 972 = 0

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D

(x + 7)^{3} - 21 (x + 7)^{2} - 400 = 0

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Solution

The correct option is A $(x - 7)^{3} - 21 (x - 7)^{2} + 972 = 0$
As $α, β, γ$ are roots of $x^{3} - 7 x + 6 = 0$
We have
$s_{1} = α + β + γ = 0 s_{2} = α β + α γ + β γ = - 7 s_{3} = α β γ = - 6$
Now
$α β + α γ + β γ = - 7 \Rightarrow α β γ + α γ^{2} + β γ^{2} = - 7 γ \Rightarrow γ^{2} (α + β) = - 7 γ - α β γ$
$\Rightarrow - γ^{3} = 6 - 7 γ \Rightarrow γ^{3} = 7 γ - 6$
Let $x = {(α - β)}^{2} = {(α + β)}^{2} - 4 α β = γ^{2} - 4 α β$
$\Rightarrow x γ = γ^{3} - 4 α β γ = 7 γ - 6 + 24 \Rightarrow γ = \frac{18}{x - 7}$
Replacing $x \to \frac{18}{x - 7}$ in given equation, we get
${(\frac{18}{x - 7})}^{3} - 7 (\frac{18}{x - 7}) + 6 = 0 \Rightarrow 18^{3} - 7 \cdot 18 (x - 7) + 6 {(x - 7)}^{3} = 0 \Rightarrow {(x - 7)}^{3} - 21 (x - 7) + 972 = 0$

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