Question

If $\alpha ,\beta ,\gamma$ are the roots of the cubic $x^3-{px}^2+qx-r=0$, find the equations whose roots are $(\beta +\gamma -\alpha ),(\gamma +\alpha -\beta ),(\alpha +\beta -\gamma )$

• A. $y^3-r{y}^2+(4r-q^2)y+p(8r-4pq+p^3)=0$
• B. $y^3-p{y}^2+(4q-p^2)y+p(8r-4pq+p^3)=0$
• C. ${ry}^3-q(q+1)y^2+{p(r+1)}^{2}y-{(p+1)}^3=0$
• D. ${ry}^3-q(r+1)y^2+{p(r+1)}^{2}y-{(r+1)}^3=0$

Answer 1

The correct option is B $y^3-p{y}^2+(4q-p^2)y+p(8r-4pq+p^3)=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3-{px}^2+qx-r=0$

Then $S_1=\alpha +\beta +\gamma =p,$

$S_2=\alpha \beta +\beta \gamma +\gamma \alpha =+q,$

$S_3=\alpha \beta \gamma =r$

Now $S_1^{\prime }=\sum (\alpha +\beta -\gamma )=\sum (p-2\gamma )$
$\therefore (\beta +\gamma -\alpha )+(\gamma +\alpha -\beta )+(\alpha +\beta -\gamma )=p$
$S_3^{\prime }=\prod (\alpha +\beta -\gamma )=(p-2\gamma )(p-2\alpha )(p-2\beta )=(p-2\gamma )(p^2-(2\alpha +2\beta )p+4\alpha \beta )=p^3-(2\alpha +2\beta )p^2+4\alpha \beta p-2{\gamma p}^2+(2\alpha +2\beta )2\gamma p-8\alpha \beta \gamma =p^3-(2\alpha +2\beta +2\gamma )p^2+p(4\alpha \beta +4\alpha \gamma +4\beta \gamma )-8\alpha \beta \gamma =p^3-{2p}^3+4pq-8r=-8r+4pq-p^3$
And
$S_2^{\prime }=(\alpha +\beta -\gamma )(\gamma +\alpha -\beta )+(\beta +\gamma -\alpha )(\gamma +\alpha -\beta )+(\beta +\gamma -\alpha )(\alpha +\beta -\gamma )=(p-2\gamma )(p-2\alpha )+(p-2\alpha )(p-2\beta )+(p-2\beta )(p-2\gamma )=p^2-(2\alpha +2\gamma )p+4\alpha \gamma +p^2-(2\alpha +2\beta )\beta +4\alpha \beta +p^2-(2\alpha +2\gamma )p+4\gamma \beta =(4q-p^2)$
Therefore using $y^3+(-S_1^{\prime })y^2+\left(S_2^{\prime }\right)y+(-S_3)=0$ required equation is

$y^3-{py}^2+(4q-p^2)y+p(8r-4pq+p^3)=0$