Question

If $A=\left|\begin{array}{ccc}a& c^2& c\\ ab& a& b\\ bc& ac& a\end{array}\right|+\left|\begin{array}{ccc}a& ab& bc\\ b& bc& c\\ c& b& a\end{array}\right|+\left|\begin{array}{ccc}{b}^2& c& b\\ b+c^2& a+bc& ac+c\\ c& b& a\end{array}\right|,$
then $A$ is divisible by

• A. $a+b+c$
• B. $a^2+b^2+c^2$
• C. $a+bw+c{w}^2,$ where $w$ be the cube root of unity
• D. $a+b{w}^2+cw,$ where $w$ be the cube root of unity

Answer 1

Answer: (A), (C), (D)

$a+b{w}^2+cw,$ where $w$ be the cube root of unity

$A=\left|\begin{array}{ccc}a& c^2& c\\ ab& a& b\\ bc& ac& a\end{array}\right|+\left|\begin{array}{ccc}a& ab& bc\\ b& bc& c\\ c& b& a\end{array}\right|+\left|\begin{array}{ccc}{b}^2& c& b\\ b+c^2& a+bc& ac+c\\ c& b& a\end{array}\right|$

$D_2\to R\leftrightarrow C,D_3\to R\leftrightarrow C$
$=\left|\begin{array}{ccc}a& c^2& c\\ ab& a& b\\ bc& ac& a\end{array}\right|+\left|\begin{array}{ccc}a& b& c\\ ab& bc& b\\ bc& c& a\end{array}\right|+\left|\begin{array}{ccc}{b}^2& b+c^2& c\\ c& a+bc& b\\ b& ac+c& a\end{array}\right|$

Add $D_1$ and $D_2$ as $C_1,C_3$ are identical.
$=\left|\begin{array}{ccc}a& b+c^2& c\\ ab& a+bc& b\\ bc& ac+c& a\end{array}\right|+\left|\begin{array}{ccc}{b}^2& b+c^2& c\\ c& a+bc& b\\ b& ac+c& a\end{array}\right|$

$=\left|\begin{array}{ccc}a+b^2& b+c^2& c\\ ab+c& a+bc& b\\ bc+b& ac+c& a\end{array}\right|$

$C_2\to C_2-c{C}_3$

$=\left|\begin{array}{ccc}a+b^2& b& c\\ ab+c& a& b\\ bc+b& c& a\end{array}\right|$

$A=a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)(a^2+w^3{b}^2+w^3{c}^2+(w+w^2)ab+(w+w^2)bc+(w+w^2\left)ca\right)=(a+b+c)(a+bw+c{w}^2)(a+b{w}^2+cw)$