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Q.
If $x \neq y \neq z$ and $\begin{vmatrix}x&x^{2}&x^{3}\\ y&y^{2}&y^{3}\\ z &z^{2}&z^{3}\end{vmatrix} = 0 $, then xyz is equal to:
Determinants
Solution:
Let $A = \begin{vmatrix}x&x^{2}&x^{3}\\ y&y^{2}&y^{3}\\ z &z^{2}&z^{3}\end{vmatrix}$
By taking x, y, z common from the rows $R_1, R_2$ and $R_3$ respectively. So,
$ A = xyz \begin{vmatrix}1&x&x^{2}\\ 1 &y&y^{2}\\ 1&z&z^{2}\end{vmatrix}$
Operate $ R_{2} \to R_{2} -R_{1} $ and $R_{3} \to R_{3} -R_{1}$
$ \Rightarrow A = xyz \begin{vmatrix}1&x&x^{2}\\ 0&y-x&y^{2}-x^{2}\\ 0&z-x&z^{2}-x^{2}\end{vmatrix} $
Now take common y - x and z - x from the rows $R_2$ and $R_3$ respectively. Thus
$ A = xyz (y - x) (z - x) \ \begin{vmatrix}1&x&x^{2}\\ 0 &1&y+x\\ 0 &1&z+x\end{vmatrix} $
= xyz (y - x) (z - x) (z - y)
= xyz (x - y) (y - z) (z - x)
Given | A | = 0
So, xyz = 0 $\because \ x \neq y \neq z$ (given)