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

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Aicharya · Accepted Answer

$\begin{aligned} & \text {Given, } \phi(x)=\frac{x}{\left(x^2+1\right)(x+1)} \ & \therefore \phi(a) \phi(b) \phi(c)=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)} \ & =\frac{a b c}{(1+a+b+c+a b+b c+c a+a b c)\left(1+a^2+b^2+c^2\right.} \ & \left.\quad+a^2 b^2+b^2 c^2+c^2 a^2+(a b c)^2\right) \end{aligned}$ Also, given that $a, b$ and $c$ are roots of cubic equation $\begin{aligned} x^3-3 x+\lambda & =0. \ a b+b c+c a & =-3 \quad \ldots (i) \ a+b+c & =0 \quad \ldots (ii) \end{aligned}$ and $a b c=-\lambda$...(iii) Squaring Eq. (ii), we get, $(a+b+c)^2=0$ $\begin{aligned} a^2+b^2+c^2+2(a b+b c+c a) & =0 \ a^2+b^2+c^2 & =6 \end{aligned}$ Similarly, $a^2 b^2+b^2 c^2+c^2 a^2=9$ (by solving) Put these values in $\phi(a) \phi(b) \phi(c)$, we get $\begin{aligned} & \frac{-\lambda}{(1+0-3-\lambda)\left(1+6+9+\lambda^2\right)} \ & \frac{-\lambda}{(-2-\lambda)\left(\lambda^2+16\right)}=\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)} \end{aligned}$

Let \(\phi(x)=\frac{x}{\left(x^2+1\right)(x+1)}\). If \(a, b\) and \(c\) are the roots of the equation…

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