If x, y, z are different and \bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3}-1 \\ y& y^{2} &y^{3}-1 \\ z&z^{2} & z^{3}-1 \end{vmatrix}= 0 ,   then using properties of determinants, show that xyz= 1\cdot

 

 

 

 
 
 
 
 

Answers (1)

\bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3}-1 \\ y& y^{2} &y^{3}-1 \\ z& z^{2} &z^{3}-1 \end{vmatrix}= 0
\Rightarrow \bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3} \\ y& y^{2} &y^{3} \\ z& z^{2} &z^{3} \end{vmatrix}+\begin{vmatrix} x & x^{2} &-1 \\ y& y^{2} &-1 \\ z& z^{2} &-1 \end{vmatrix}= 0
Taking x, y, z common from R1,R and R3  respectively in the first determinant . Also taking -1 common from C3 be the second determinant
\Rightarrow xyz\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}-\begin{vmatrix} x & x^{2}&1 \\ y& y^{2}&1 \\ z& z^{2} &1 \end{vmatrix}= 0
In second determinant  C1\leftrightarrow C3
\Rightarrow xyz\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}-\begin{vmatrix} 1 & x^{2}&x \\ 1& y^{2}&y \\ 1& z^{2} &z \end{vmatrix}= 0
In second determinant  C2 \leftrightarrow C3
\Rightarrow xyz\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}-\begin{vmatrix} 1 & x&x^{2} \\ 1& y&y ^{2}\\ 1& z &z^{2} \end{vmatrix}= 0
\Rightarrow \left ( xyz-1 \right )\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}= 0
By R_{1}\rightarrow R_{1}-R_{3}\; \; ,R_{2}\rightarrow R_{2}-R_{3}
\Rightarrow \left ( xyz-1 \right )\begin{vmatrix} 0 & x-z &x^{2}-z^{2} \\ 0 &y-z &y^{2}-z^{2} \\ 1 &z &z^{2} \end{vmatrix}= 0
Taking \left ( x-z \right )\; and\; \left ( y-z \right )  common from  R1 and R2 respectively.
\Rightarrow \left ( xyz-1 \right )\left ( y-z \right )\left ( x-z \right )\begin{vmatrix} 0 &1 &x+z \\ 0& 1 &y+z \\ 1&z & z^{2} \end{vmatrix}= 0

Expanding along C1
\Rightarrow \left ( xyz-1 \right )\left ( y-z \right )\left ( x-z \right )\left ( y+z-z-x \right )= 0
\Rightarrow \left ( xyz-1 \right )\left ( y-z \right )\left ( x-z \right )\left ( y-x \right )= 0
AS x\neq y\neq z\; \; \therefore \left ( xyz-1 \right )= 0\Rightarrow xyz= 1
 \therefore xyz= 1

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