Question

If $a$, $b$, $c$ are the roots of the equation $x^3-3x^2+3x+7=0$, then the value of $\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|$ is

• A. $9$
• B. $27$
• C. $81$
• D. $0$

Answer 1

Answer: (D)

$x^3-3x^2+3x+7=0$
$\Rightarrow {\left(x-1\right)}^3+8=0\Rightarrow {\left(x-1\right)}^3={\left(-2\right)}^3$
$\Rightarrow {\left(\frac{x-1}{-2}\right)}^3=1\Rightarrow \frac{x-1}{2}={\left(1\right)}^{13}=1,\omega ,{\omega }^2$
$\Rightarrow x-1=-2,-2\omega ,-2{\omega }^2\Rightarrow x=-1,1-2\omega ,1-2{\omega }^2$
$\Rightarrow a=-1,b=1-2\omega ,c=1-2{\omega }^2$
Now, $\Delta =\left|\begin{array}{ccc}2bc-a^2& c^2& b^2\\ c^2& 2ac-b^2& a^2\\ b^2& a^2& 2ab-c^2\end{array}\right|={\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^2$
$={\left[-\left(a^3+b^3+c^3-3abc\right)\right]}^2$
$={\left\{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right\}}^2$
$\frac{1}{4}{\left(a+b+c\right)}^2{\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\}}^2=0$