Question

If $a^3\left|\begin{array}{ccc}1& b& c\\ b^2& b^3+1& b^{2}c\\ c^2& b{c}^2& c^3+1\end{array}\right|+a^2\left|\begin{array}{ccc}\frac{1}{a^2}& 0& 0\\ b& b^3+1& b{c}^2\\ c& b^{2}c& c^3+1\end{array}\right|=18,$
where $a,b,c\ge 0,$ then the values of $(a,b,c)$ satisfying the given equation is/are

• A. $(1,2,2)$
• B. $(2,1,2)$
• C. $(2,1,1)$
• D. $(1,1,2)$

Answer 1

Answer: (A), (B)

The correct options are
A $(1,2,2)$
B $(2,1,2)$

$a^3\left|\begin{array}{ccc}1& b& c\\ b^2& b^3+1& b^{2}c\\ c^2& b{c}^2& c^3+1\end{array}\right|+a^2\left|\begin{array}{ccc}\frac{1}{a^2}& 0& 0\\ b& b^3+1& b{c}^2\\ c& b^{2}c& c^3+1\end{array}\right|=18$
Multiplying $a^2$ in $R_1$ and $a$ in $C_1$ to first determinant and $R\leftrightarrow C$ in second determinant then multiply $a^2$ in $R_1$
$\left|\begin{array}{ccc}{a}^3& a^{2}b& a^{2}c\\ a{b}^2& b^3+1& b^{2}c\\ a{c}^2& b{c}^2& c^3+1\end{array}\right|+\left|\begin{array}{ccc}1& a^{2}b& a^{2}c\\ 0& b^3+1& b^{2}c\\ 0& b{c}^2& c^3+1\end{array}\right|=18\Rightarrow \left|\begin{array}{ccc}{a}^3+1& a^{2}b& a^{2}c\\ a{b}^2& b^3+1& b^{2}c\\ a{c}^2& b{c}^2& c^3+1\end{array}\right|=18R_1\to a{R}_1,R_2\to b{R}_2,R_3\to c{R}_3\Rightarrow \left|\begin{array}{ccc}{a}^4+a& a^{3}b& a^{3}c\\ a{b}^3& b^4+b& b^{3}c\\ a{c}^3& b{c}^3& c^4+c\end{array}\right|=18abc\Rightarrow abc\left|\begin{array}{ccc}{a}^3+1& a^3& a^3\\ b^3& b^3+1& b^3\\ c^3& c^3& c^3+1\end{array}\right|=18abc{R}_1\to R_1+R_2+R_3(1+a^3+b^3+c^3)\left|\begin{array}{ccc}1& 1& 1\\ b^3& b^3+1& b^3\\ c^3& c^3& c^3+1\end{array}\right|=18C_2\to C_2-C_1,C_3\to C_3-C_1(1+a^3+b^3+c^3)\left|\begin{array}{ccc}1& 0& 0\\ b^3& 1& 0\\ c^3& 0& 1\end{array}\right|=18\Rightarrow 1+a^3+b^3+c^3=18\Rightarrow a^3+b^3+c^3=17\therefore (a,b,c)=(1,2,2),(2,1,2),(2,2,1)$