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If $$f(x)=a+bx+{ cx }^{ 2 }$$ and $$\alpha ,\beta ,\gamma $$ are the roots of the equation $${ x }^{ 3 }=1$$, then $$\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right| $$ is equal to


A
f(α)+f(β)+f(γ)
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B
f(α)f(β)+f(β)f(γ)+f(γ)f(α)
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C
f(α)f(β)f(γ)
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D
f(α)f(β)f(γ)
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Solution

The correct option is D $$-f(\alpha )f(\beta )f(\gamma )$$
$$f(x)=a+bx+{ cx }^{ 2 }$$ and $$\alpha ,\beta ,\gamma $$ are the roots of the equation $${ x }^{ 3 }=1$$
i.e $$\alpha=1,\beta=\omega,\gamma=\omega^2$$ since $$1,\omega,\omega^2$$ are cube roots of unity.
$$\therefore \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$$
$$=(a+b+c)\begin{vmatrix} 1 & 1 &1  \\ b & c & a \\ c & a & b \end{vmatrix}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$=-(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+\omega^4)$$
$$=-f(\alpha)f(\beta)f(\gamma)$$
Hence, option D.

Mathematics

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