Question

The number of positive integral solutions of the equation $\left|\begin{array}{ccc}{x}^3+1& x^{2}y& x^{2}z\\ x{y}^2& y^3+1& y^{2}z\\ x{z}^2& y{z}^2& z^3+1\end{array}\right|=11$ is/are

• A. $1$
• B. $3$
• C. $6$
• D. $12$

Answer 1

Answer: (B)

Multiplying $R_1$ by $x$ , $R_2$ by $y$ and $R_3$ by $z$ and dividing the determinant by $xyz$ , we get,
$\frac{1}{xyz}\left|\begin{array}{ccc}{x}^4+x& x^{3}y& x^{3}z\\ x{y}^3& y^4+y& y^{3}z\\ x{z}^3& y{z}^3& z^4+z\end{array}\right|=11$
$\Rightarrow \frac{xyz}{xyz}\left|\begin{array}{ccc}{x}^3+1& x^3& x^3\\ y^3& y^3+1& y^3\\ z^3& z^3& z^3+1\end{array}\right|=11$
Use $R_1\to R_1+R_2+R_3$
$\left(x^3+y^3+z^3+1\right)\left|\begin{array}{ccc}1& 1& 1\\ y^3& y^3+1& y^3\\ z^3& z^3& z^3+1\end{array}\right|=11$
$\Rightarrow x^3+y^3+z^3=10$ (as the value of the determinant is $1$ )
Hence, all the possible solutions are $\left(2,1,1\right),\left(1,2,1\right),\left(1,1,2\right)$