Question

If $\alpha ,\beta and\gamma$ are the roots of the equation $x^3+3x+2=0$ , Find the equation whose roots are $\alpha -\beta \left)\right(\alpha -\beta ),(\beta -\gamma \left)\right(\beta -\alpha ),(\gamma -\alpha \left)\right(\gamma -\beta$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\alpha ,\beta and\gamma$ are the roots of the equation $x^3+3x+2=0$ ________(1)

Sum of the roots $\alpha +\beta +\gamma =\frac{-b}{a}=0\to \beta +\gamma =-\alpha$

Sum of the roots taking two at a time $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=3$

Product of the roots $\alpha \gamma \beta =\frac{-d}{a}=-2\to \beta \gamma =\frac{-2}{\alpha }$

Let $y=(\alpha -\beta )(\alpha -\gamma )$

$y=(\alpha -\beta )(\alpha -\gamma )={\alpha }^2-\alpha \beta -\alpha +\beta$

= ${\alpha }^2-\alpha (\beta +\gamma )+\beta \gamma \beta +\gamma =-\alpha ,\beta \gamma =\frac{-2}{\alpha }$

$={\alpha }^2-\alpha (-\alpha )+\frac{-2}{\alpha }$

$y=2{\alpha }^2-\frac{2}{\alpha }$

$y\alpha =2{\alpha }^3-2$ ___________(2)

$2{\alpha }^3-y\alpha -2=0$

To generalize this equation

Replace $\alpha =x$

2$x^3$ -yx - 2 = 0 ____________(3)

To get the relation between x and y

Subtracting equation 3 from twice of equation 1

2$x^3$ -xy - 2 - 2$x^3$ - 6x - 4 = 0

-xy - 6x - 6 = 0

x(6 + y) = - 6

Now replace

x = $\frac{-6}{6+y}$ in equation 1

-$\frac{216}{{(6+y)}^3}$ - $\frac{18}{6+y}$ + 2 = 0

${(y+6)}^3-9{(y+6)}^2-108=0$

$y^3+9y^2-216=0$

Replace y by x $\Rightarrow x^3+9x^3-216=0$

$x^3+9x^2-216=0$ has roots

$(\alpha -\beta )(\alpha -\gamma ),(\beta -\gamma )(\beta -\alpha ),(\gamma -\alpha )(\gamma -\beta )$