Question

If abc $\ne$0 and if $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ = 0 then $\frac{a^3+b^3+c^3}{abc}$

• A. 3
• B. -3
• C. 2
• D. -2

Answer 1

The correct option is A 3

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
$=a(bc-a^2)-b(b^2-ac)+c(ab-c^2)$
$=3abc-a^3-b^3-c^3$
Given,
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

So, $\frac{a^3+b^3+c^3}{abc}=3$