Question

If $a,b,c$ are roots of the equation $x^3+p{x}^2+q=0$, then the value of the determinant $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$, is equal to

• A. $p^3$
• B. $p^3-3q$
• C. $-p^3$
• D. $p^2-3q$

Answer 1

Answer: (A)

$p^3$

As $a,b,c$ are roots of $x^3+p{x}^2+0x+q=0$

Then

$s_1=a+b+c=-p$

$s_2=ab+bc+ca=0$

$s_3=abc=-q$

Now,

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|$

$=-(a+b+c)[a^2+b^2+c^2-ab-bc-ca]$

$=-(a+b+c)[{(a+b+c)}^2-3(ab+bc+ca)]$

$=-{(a+b+c)}^3=-{(-p)}^3=p^3$

Hence, option 'A' is correct.