\(\begin{aligned}
& \text {Given, } \phi(x)=\frac{x}{\left(x^2+1\right)(x+1)} \\
& \therefore \phi(a) \phi(b) \phi(c)=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)} \\
& =\frac{a b c}{(1+a+b+c+a b+b c+c a+a b c)\left(1+a^2+b^2+c^2\right.} \\
& \left.\quad+a^2 b^2+b^2 c^2+c^2 a^2+(a b c)^2\right)
\end{aligned}\)
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
\(\begin{aligned}
& \frac{-\lambda}{(1+0-3-\lambda)\left(1+6+9+\lambda^2\right)} \\
& \frac{-\lambda}{(-2-\lambda)\left(\lambda^2+16\right)}=\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}
\end{aligned}\)