Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\begin{array}{c}\Rightarrow \left|\begin{array}{ccc}{y}^3+1& y^{2}z& y^{2}x\\ y{z}^2& x^3+1& z^{2}x\\ y{x}^2& x^{2}z& x^3+1\end{array}\right|=11\to \left(i\right)\\ Takingouty,z,xin{c}_1,c_2,c_3.weget\\ xyz\left|\begin{array}{ccc}\frac{y^3+1}{y}& y^2& y^2\\ z^2& \frac{z^3+1}{z}& z^2\\ x^2& x^2& \frac{x^3+1}{x}\end{array}\right|\\ Multiplyy,z\&xin{R}_1,R_2,\&R_3,weget\\ \Rightarrow \left|\begin{array}{ccc}{y}^3+1& y^3& y^3\\ z^3& z^3+1& z^3\\ x^3& x^3& x^3+1\end{array}\right|\\ Apply{R}_1\to R_1+R_2+R_3\\ \Rightarrow \left|\begin{array}{ccc}{x}^3+y^3+z^3+1& x^3+y^3+z^3+1& x^3+y^3+z^3+1\\ z^3& z^3+1& z^3\\ x^3& x^3& x^3+1\end{array}\right|\\ Takingoutcommon\left(x^3+y^3+z^3+1\right)from{R}_1,wege{t}^{\prime }\\ \left(x^3+y^3+z^3+1\right)\left|\begin{array}{ccc}1& 1& 1\\ z^3& z^3+1& z^3\\ x^3& x^3& x^3+1\end{array}\right|Apply{C}_2\to c_2-c_1\&c_3\to c_3-c_1\\ \Rightarrow \left(x^3+y^3+z^3+1\right)\left|\begin{array}{ccc}1& 0& 0\\ z^3& 1& 0\\ x^3& 0& 1\end{array}\right|\Rightarrow \left(x^3+y^3+z^3+1\right)\mathrm{1.1.1}\\ Nowfromequation{x}^3+y^3+z^3+1=11\Rightarrow x^3+y^3+z^3=10\end{array}$

According to hit and trial method, we get

$\begin{array}{cc}Putx=2,y=1,z=1& Similarly,\left(1,2,1\right)\&\left(1,1,2\right)\\ 2^3+1^3+1^3=10& So,\left(2,1,1\right),\left(1,2,1\right)\&\left(1,1,2\right)\\ 10=10& \end{array}$