Question

If $\alpha ,\beta$ and $\gamma$ are the roots of the equation $x^3+2x^2+3x+1=0$. Find the constant term of the equation whose roots are $\frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}-\frac{1}{{\alpha }^3},\frac{1}{{\gamma }^3}+\frac{1}{{\alpha }^3}-\frac{1}{{\beta }^3},\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}-\frac{1}{{\gamma }^3}$.

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Answer 1

$x^3+2x^2+3x+1=0$

$x^3+1=-2x^2+3x$ . . . $\left(1\right)$

Cubing on both sides

${(x^3+1)}^3=-{(2x^2+3x)}^3$

$\Rightarrow x^9+3x^6+3x^3+1=-[8x^6+27x^3+18x^3(2x^2+3x\left)\right]$

$\Rightarrow x^3+1=-2x^2+3x$ from equation $\left(1\right)$.

$\Rightarrow x^9+3x^6+3x^3+1=-[8x^6+27x^3+18x^3(-x^3-1\left)\right]$

Putting $x^3=y$ in this equation

$\Rightarrow y^3+3y^2+3y+1=-[8y^2+27y+18y(-y-1\left)\right]$

$y^3-7y^2+12y+1=0$ (its root ${\alpha }^3,{\beta }^3,{\gamma }^3$) . . . $\left(2\right)$

Changing $y$ to $\frac{1}{y}$ in equation $2$

$\Rightarrow \frac{1}{y^3}-\frac{7}{y^2}+\frac{12}{y}+1=0$

$\Rightarrow y^3+12y^2-7y+1=0$ . . . $\left(3\right)$

If roots are $\frac{1}{{\alpha }^3},\frac{1}{{\beta }^3},\frac{1}{{\gamma }^3}$.

Let $\frac{1}{{\alpha }^3}=a,\frac{1}{{\beta }^3}=b,\frac{1}{{\gamma }^3}=c$

$\Rightarrow a+b+c=-12$

$\Rightarrow \frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}-\frac{1}{{\alpha }^3}=(a+b+c)-2a=-12-2y$

$\Rightarrow y=\frac{12+p}{2}$

Putting the value of $y$ in equation $3$

$\Rightarrow -\frac{(12+p{)}^3}{8}+12\frac{(12+p{)}^2}{4}+7\frac{(12+p)}{2}+1=0$

$\Rightarrow (12+p{)}^3-24(12+p{)}^2-28(12+p)-8=0$

$\Rightarrow p^3+12p^2-172p-2072=0$

Change the variable $x$ in place of $p$,

$\Rightarrow x^3+12x^2-172x-2072$

Constant term $=-2072$