If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3 x^2+4 x$ $+5=0$, then the cubic…

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3 x^2+4 x$ $+5=0$, then the cubic equation whose roots are $1+4 \alpha$, $1+4 \beta$ and $1+4 \gamma$ is
  1. $x^3+9 x^2-21 x+267=0$
  2. $x^3+9 x^2+43 x+267=0$
  3. $x^3+9 x^2+41 x+267=0$
  4. $x^3+9 x^2+43 x+267=0$

Solution

$\alpha, \beta, \gamma$ 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} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

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