Let phi(x)=frac{x}{(x^2+1)(x+1)}. If a, b and | vedclass

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(D) Given $\phi(x) = \frac{x}{(x^2+1)(x+1)}$.We need to find $\phi(a) \phi(b) \phi(c) = \frac{abc}{(a^2+1)(b^2+1)(c^2+1)(a+1)(b+1)(c+1)}$.For the equa

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$\frac{\lambda}{(\lambda+2)(\lambda^2+16)}$

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(D) Given $\phi(x) = \frac{x}{(x^2+1)(x+1)}$.We need to find $\phi(a) \phi(b) \phi(c) = \frac{abc}{(a^2+1)(b^2+1)(c^2+1)(a+1)(b+1)(c+1)}$.For the equation $x^3 - 3x + \lambda = 0$,we have:$a+b+c = 0$,$ab+bc+ca = -3$,and $abc = -\lambda$.First,$(a+1)(b+1)(c+1) = abc + (ab+bc+ca) + (a+b+c) + 1 = -\lambda - 3 + 0 + 1 = -\lambda - 2 = -(\lambda+2)$.Second,$(a^2+1)(b^2+1)(c^2+1) = (a^2b^2+a^2+b^2+1)(c^2+1) = a^2b^2c^2 + a^2b^2 + a^2c^2 + a^2 + b^2c^2 + b^2 + c^2 + 1$.Using $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 0^2 - 2(-3) = 6$.Using $a^2b^2+b^2c^2+c^2a^2 = (ab+bc+ca)^2 - 2abc(a+b+c) = (-3)^2 - 2(-\lambda)(0) = 9$.So,$(a^2+1)(b^2+1)(c^2+1) = (abc)^2 + (a^2b^2+b^2c^2+c^2a^2) + (a^2+b^2+c^2) + 1 = \lambda^2 + 9 + 6 + 1 = \lambda^2 + 16$.Thus,$\phi(a) \phi(b) \phi(c) = \frac{-\lambda}{(\lambda^2+16)(-(\lambda+2))} = \frac{\lambda}{(\lambda+2)(\lambda^2+16)}$.

Let $\phi(x)=\frac{x}{(x^2+1)(x+1)}$. If $a, b$ and $c$ are the roots of the equation $x^3-3x+\lambda=0, (\lambda \neq 0)$,then $\phi(a) \phi(b) \phi(c) =$

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