Question

If $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+c=0;a,b,c\in I^{+}$, such that $\Delta =\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$, then $\Delta$ is always divisible by

• A. $(a+b+c{)}^2$
• B. $b^2-4ac$
• C. $\frac{1}{a^4}$
• D. $a^4$

Answer 1

The correct option is C $\frac{1}{a^4}$

$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$

$\Delta =\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$

Splitting it as a product of two determinants.
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
$={\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2$

$C_1\to C_1-C_2,C_2\to C_2-C_3$
$={\left|\begin{array}{ccc}0& 0& 1\\ 1-\alpha & \alpha -\beta & \beta \\ 1-{\alpha }^2& {\alpha }^2-{\beta }^2& {\beta }^2\end{array}\right|}^2$
$=\left[\right(1-\alpha \left)\right(a-\beta ){]}^2{\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& \beta \\ 1+\alpha & \alpha +\beta & {\beta }^2\end{array}\right|}^2$
$=\left[\right(1-\alpha \left)\right(\alpha -\beta \left){]}^2\right(\beta -1{)}^2$
$=\left[\right(1-\alpha \left)\right(1-\beta \left){]}^2\right(\alpha -\beta {)}^2$
$=[1-(\alpha +\beta )+\alpha \beta {]}^2[(\alpha +\beta {)}^2-4\alpha \beta ]$
$={[1+\frac{b}{a}+\frac{c}{a}]}^2[\frac{b^2}{a^2}-\frac{4c}{a}]$
$=\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$