Question

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

• A. $9$
• B. $27$
• C. $81$
• D. $0$

Answer 1

Answer: (D)

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