Question

The number of positive integral solutions of the equation $\begin{vmatrix} x^{3}+1 & x^{2}y & x^{2}z \\ xy^{2} & y^{3}+1 & y^{2}z \\ xz^{2} & yz^{2} & z^{3}+1 \end{vmatrix}=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,
$\frac{1}{x y z}\begin{vmatrix} x^{4}+x & x^{3}y & x^{3}z \\ xy^{3} & y^{4}+y & y^{3}z \\ xz^{3} & yz^{3} & z^{4}+z \end{vmatrix}=11$
$\Rightarrow \frac{x y z}{x y z}\begin{vmatrix} x^{3}+1 & x^{3} & x^{3} \\ y^{3} & y^{3}+1 & y^{3} \\ z^{3} & z^{3} & z^{3}+1 \end{vmatrix}=11$
Use $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$
$\left(x^{3} + y^{3} + z^{3} + 1\right)\begin{vmatrix} 1 & 1 & 1 \\ y^{3} & \, y^{3}+1 & y^{3} \\ z^{3} & z^{3} & z^{3}+1 \end{vmatrix}=11$
$\Rightarrow x^{3}+y^{3}+z^{3}=10$ (as the value of the determinant is $1$ )
Hence, all the possible solutions are $\left(2 , 1 , 1\right),\left(1 , 2 , 1\right),\left(1 , 1 , 2\right)$