Question

If $\alpha ,\beta ,\gamma$ are the roots of the cubic $x^3+qx+r=0$, then the value of $\Pi (\alpha -\beta {)}^2=$

• A. $-(27q^2+4r^3)$
• B. $-(27q+4r^2)$
• C. $-(27r^2+4q^3)$
• D. $-(27r+4q^2)$

Answer 1

The correct option is C $-(27r^2+4q^3)$

Given that,

$\alpha ,\beta ,\gamma ,$ are the roots of the cubic equation $x^3+qx+r=0$

So,

$\alpha +\beta +\gamma =-b/a=0$

$\alpha \beta +\beta \gamma +\gamma \alpha =c/a=q$

$\alpha \beta \gamma =-d/a=-r$

$\prod (\alpha -\beta {)}^2$

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

$=\left[\right(\alpha -\beta \left)\right(\beta -\gamma \left)\right(\gamma -\alpha ){]}^2$

$=[{\alpha }^2\gamma -{\alpha }^2\beta +{\beta }^2\alpha -{\beta }^2\gamma +{\gamma }^2\beta -{\gamma }^2\alpha {]}^2$

$=\left[\alpha \right(\alpha \gamma -\alpha \beta )+\beta (\beta \alpha -\beta \gamma )+\gamma (\gamma \beta -\gamma \alpha ){]}^2$

$=\left[\alpha \right(2\alpha \gamma +\gamma \alpha -q)+\beta (2\alpha \beta +\gamma \alpha -q)+\gamma (2\beta \gamma +\alpha \beta -q){]}^2$ $(\because \alpha \beta +\beta \gamma +\gamma \alpha =q)$

$=\left[2\right(\alpha {\beta }^2+\beta {\gamma }^2+\gamma {\alpha }^2)+3\alpha \beta \gamma -q(\alpha +\beta +\gamma ){]}^2$

$=[3\alpha \beta \gamma +2\alpha \beta \gamma (\frac{\beta }{\alpha }+\frac{\gamma }{\alpha }+\frac{\alpha }{\beta })+0{]}^2$ $(\because \alpha +\beta +\gamma =0)$

Another Approach: by value putting

Let, $r=0andq=-1$

Then the cubic equation becomes :

$x^3-x=0$

$⟹x(x+1)(x-1)=0$

$⟹x=0,1,-1$

So, $\alpha =0,\beta =1,\gamma =-1$

$\therefore \prod (\alpha -\beta {)}^2$

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

$=4$

For $q=-1andr=0$

only option C = 4

Hence, option C is correct.