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(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2+4x+5=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -3$ $\alpha\beta+\beta\gamma+\gamma\al

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$x^3+9x^2+43x+267=0$

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(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2+4x+5=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -3$ $\alpha\beta+\beta\gamma+\gamma\alpha = 4$ $\alpha\beta\gamma = -5$ Let $A=1+4\alpha, B=1+4\beta, C=1+4\gamma$. Sum of roots: $A+B+C = 3+4(\alpha+\beta+\gamma) = 3+4(-3) = -9$. Sum of roots taken two at a time: $AB+BC+CA = (1+4\alpha)(1+4\beta) + (1+4\beta)(1+4\gamma) + (1+4\gamma)(1+4\alpha) = 3+8(\alpha+\beta+\gamma) + 16(\alpha\beta+\beta\gamma+\gamma\alpha) = 3+8(-3)+16(4) = 3-24+64 = 43$. Product of roots: $ABC = (1+4\alpha)(1+4\beta)(1+4\gamma) = 1+4(\alpha+\beta+\gamma)+16(\alpha\beta+\beta\gamma+\gamma\alpha)+64(\alpha\beta\gamma) = 1+4(-3)+16(4)+64(-5) = 1-12+64-320 = -267$. The required cubic equation is $x^3 - (A+B+C)x^2 + (AB+BC+CA)x - ABC = 0$. Substituting the values: $x^3 - (-9)x^2 + 43x - (-267) = 0$,which simplifies to $x^3+9x^2+43x+267=0$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3x^2+4x+5=0$,then the cubic equation whose roots are $1+4\alpha$,$1+4\beta$ and $1+4\gamma$ is

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