$\alpha, \beta, \gamma$ are roots of $x^3+3 x^2+4 x+5=0$
$\alpha+\beta+\gamma=-3$ ...(i)
$\alpha \beta+\beta \gamma+\gamma \alpha=4$ ...(ii)
$\alpha \beta \gamma=-5$ ...(iii)
$$
\begin{aligned}
& \text { Let } A=1+4 \alpha, B=1+4 \beta, C=1+4 \gamma \\
& \text { then }(A+B+C)=(1+4 \alpha)+(1+4 \beta)+(1+4 \gamma) \\
& =3+4(\alpha+\beta+\gamma)=3+4(-3)=-9 \\
& \text { and } A B+B C+C A=(1+4 \alpha)(1+4 \beta)+(1+4 \beta)(1+4 \gamma) \\
& +(1+4 \gamma)(1+4 \alpha) \\
& =1+4 \beta+4 \alpha+16 \alpha \beta+1+4 \beta+4 \gamma+16 \beta \gamma+1+4 \gamma \\
& +4 \alpha+16 \gamma \alpha \\
& =3+4(2 \beta+2 \alpha+2 \gamma)+16(\alpha \beta+\beta \gamma+\gamma \alpha) \\
& =3+8(\alpha+\beta+\gamma)+16(\alpha \beta+\beta \gamma+\gamma \alpha) \\
& =3+8(-3)+16(4)=43
\end{aligned}
$$
$$
\begin{aligned}
& \text { and } A B C=(1+4 \alpha)(1+4 \beta)(1+4 \gamma) \\
& =(1+4 \alpha+4 \beta+16 \alpha \beta)(1+4 \gamma) \\
& =(1+4 \alpha+4 \beta+16 \alpha \beta+4 \gamma+16 \alpha \gamma+16 \beta \gamma+64 \alpha \beta \gamma) \\
& =1+4(\alpha+\beta+\gamma)+16(\alpha \beta+\beta \gamma+\gamma \alpha)+64(\alpha \beta \gamma) \\
& =1+4(-3)+16(4)+64(-5)=-267
\end{aligned}
$$
therefore the require cubic equation with roots $A, B$ and $C$ is
$$
\begin{aligned}
& x^3-(A+B+C) x^2+(A B+B C+C A) x-A B C=0 \\
& \Rightarrow x^3-(-9) x^2+(43) x-(-267)=0 \\
& \Rightarrow x^3+9 x^2+43 x+267=0
\end{aligned}
$$