Question

The number of triplets $(a,b,c)$ of positive integers satisfying the equation $\left|\begin{array}{ccc}{a}^3+1& a^{2}b& a^{2}c\\ a{b}^2& b^3+1& b^{2}c\\ a{c}^2& b{c}^2& c^3+1\end{array}\right|=11$ is equal to

• A. $0$
• B. $3$
• C. $6$
• D. $12$

Answer 1

The correct option is B $3$

$\left|\begin{array}{ccc}{a}^3+1& a^{2}b& a^{2}c\\ a{b}^2& b^3+1& b^{2}c\\ a{c}^2& b{c}^2& c^3+1\end{array}\right|=11$
$\Rightarrow \frac{1}{abc}\left|\begin{array}{ccc}a(a^3+1)& a^{3}b& a^{3}c\\ a{b}^3& b(b^3+1)& b^{3}c\\ a{c}^3& b{c}^3& c(c^3+1)\end{array}\right|=11$

$\Rightarrow \frac{abc}{abc}\left|\begin{array}{ccc}{a}^3+1& a^3& a^3\\ b^3& b^3+1& b^3\\ c^3& c^3& c^3+1\end{array}\right|=11$

Applying $C_1\to C_1-C_3$ and $C_2\to C_2-C_3,$ we get
$\left|\begin{array}{ccc}1& 0& a^3\\ 0& 1& b^3\\ -1& -1& c^3+1\end{array}\right|=11$

$\Rightarrow a^3+b^3+c^3+1=11$
$\Rightarrow a^3+b^3+c^3=10$
$\therefore$ Possible triplets are $(1,1,2),$ $(1,2,1)$ and $(2,1,1)$
Hence, number of triplets $=3$