Question

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

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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