Question

If $\Delta =\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$, then the value of $\left|\begin{array}{ccc}{a}^2-bc& b^2-ca& c^2-ab\\ c^2-ab& a^2-bc& b^2-ca\\ b^2-ca& c^2-ab& a^2-bc\end{array}\right|$ is

• A. ${\Delta }^2$
• B. $2{\Delta }^2$
• C. ${\Delta }^3$
• D. none of these

Answer 1

The correct option is A ${\Delta }^2$

$\Delta =\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|=a(a^2-bc)-b(-b^2+ac)+c(c^2-ab)=a^3-abc-abc+b^3+c^3-abc=a^3+b^3+c^3-3abc$
Now,
$\left|\begin{array}{ccc}{a}^2-bc& b^2-ac& c^2-ab\\ c^2-ab& a^2-bc& b^2-ac\\ b^2-ac& c^2-ab& a^2-bc\end{array}\right|=\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ c^2& a^2& b^2\\ b^2& c^2& a^2\end{array}\right|-\left|\begin{array}{ccc}bc& ca& ab\\ ab& bc& ac\\ ac& ab& bc\end{array}\right|=a^2(a^4-b^2{c}^2)-b^2({-b}^4+a^2{c}^2)+c^2(c^4-b^2{a}^2)-\{bc(b^2{c}^2-a^{2}bc)-ac(ac{b}^2-a^2{c}^2)+ab(b^2{a}^2-ab{c}^2)\}={(a^3+b^3+c^3-3abc)}^2={\Delta }^2$